On the determination of the number of positive and negative polynomial zeros and their isolation

Emil M. Prodanov

doi:10.1515/math-2020-0079

Article Open Access

On the determination of the number of positive and negative polynomial zeros and their isolation

Emil M. Prodanov

Published/Copyright: December 1, 2020

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$Open Mathematics$

From the journal Open Mathematics Volume 18 Issue 1

Abstract

A novel method with two variations is proposed with which the number of positive and negative zeros of a polynomial with real coefficients and degree n can be restricted with significantly better determinacy than that provided by the Descartes rule of signs. The method also allows the isolation of the zeros of the polynomial quite successfully, and the determined root bounds are significantly narrower than the Cauchy and the Lagrange bounds. The method relies on solving equations of degree smaller than that of the given polynomial. One can determine analytically the exact number of positive and negative zeros of a polynomial of degree up to and including five and also fully isolate the zeros of the polynomial analytically, and with one of the variations of the method, one can analytically approach polynomials of degree up to and including nine by solving equations of degree not more than four. For polynomials of higher degree, either of the two variations of the method should be applied recursively. Numerous examples are given. Presented is the full classification of the roots of the cubic equation, together with their isolation intervals – especially important for various scientific models for which the coefficients of the equation might be functions of the model parameters. An application of the method to a quartic equation with variable coefficients (resulting from the study of wave–current interactions in the physically realistic model of azimuthal two-dimensional non-viscous flows with piecewise constant vorticity in a two-layer fluid with a flat bed and a free surface) has been demonstrated. A step-by-step algorithm for each version of the method has also been presented.

Keywords: roots of equations; isolation intervals; Descartes rule

MSC 2010: 11D41; 26C10

1 Introduction

According to the Descartes’ rule of signs [1] from 1637, an algebraic equation with real coefficients cannot have more positive real roots than the variations of sign in the sequence of its coefficients. Gauss showed [2] in 1876 that the number of positive real roots (counted with their multiplicity) is, more precisely, either equal to the number of variations of signs in the sequence of the coefficients or is equal to the number of variations of signs in the sequence of the coefficients reduced by an even number.

Many extensions of the Descartes rule have been proposed – see [3], in which Marden has given a thorough summary of the results about polynomial roots. For example, if I = ( a , b ) is an arbitrary open interval, the mapping x ↦ ( a x + b ) / ( x + 1 ) maps ( 0 , ∞ ) bijectively onto ( a , b ) . Hence, if p n ( x ) is a polynomial of degree n, then the positive real zeros of ( 1 + x ) n p n [ ( a x + b ) / ( x + 1 ) ] correspond bijectively to the real zeros of p n ( x ) in I [4].

Bounds on the zeros of polynomials were first presented by Lagrange [5] and Cauchy [6]. Ever since, the determination of the number of positive and negative roots of an equation together with finding root bounds has been the subject of intensive research. A more recent survey is provided by Pan [7]. Currently, the best root isolation techniques are subdivision methods with Descartes’ rule terminating the recursion.

In this work, a novel method is proposed for the determination of the number of positive and negative zeros of a given polynomial p n ( x ) with real coefficients and of degree n. The method also allows finding the bounds on the zeros of p n ( x ) . These bounds will not be on all of the zeros as a bulk, but, rather, if the bounds are not found individually, thus isolating each of the zeros, then not many of the zeros of the polynomial would be clumped into one isolation interval. All of this is achieved by considering the given polynomial p n ( x ) as a difference of two polynomials, the intersections of whose graphs give the roots of p n ( x ) = 0 . The idea of the method is to extract information about the roots of the given polynomial by solving equations of degree lower than n – in some sense, by “decomposing” p n ( x ) into its ingredients and studying the interaction between them. Different decompositions (further referred to as “splits”) yield different perspectives. Two splits are studied and illustrated in detail with numerous examples (with a different approach to one of the splits mentioned at the end). The first variation of the method splits the polynomial p n ( x ) by presenting it as a difference of two polynomials one of which is of degree n, but for which the origin is a zero of order k < n , while the other polynomial in the split is of degree k − 1 . For instance, polynomials of degree 9 can be split in the “middle” with k = 5 in which case all resulting equations that need to be solved are of degree 4 and their roots can be found analytically. This proves to be a very rich source of information about the zeros of this given polynomial of degree 9. This is illustrated with an example in which the negative zeros, together with one of the positive zeros of the example polynomial, have all been isolated, while the remaining two positive zeros are found to be within two bounds. All of the respective bounds are well within the bulk bounds, as found using the Lagrange and Cauchy formulae. The advantages over the Descartes rule of signs are also clearly demonstrated. If the degree of the polynomial is higher, then the method should be applied recursively.

The second variation of the method splits p n ( x ) by selecting the very first and the very last of its terms and “propping” them against the remaining ones. This split is studied in minute detail and the full classification of the roots of the cubic equation is presented, together with their isolation intervals and the criterion for the classification. The idea again relies on the “interaction” between two, this time different “ingredients” of p n ( x ) . One of these is a curve the graph of which passes through point ( 0 , 1 ) , while the graph of the other one passes through the origin. By varying the only coefficient of the former and by solving equations of degree n − 1 , one can easily find the values which would render the two graphs tangent to each other and also find the points at which this happens. Then simple comparison of the given coefficient of the leading term to these values immediately determines the exact number of positive and negative roots and their isolation intervals for any equation of degree 5 or less. For polynomials of degree 6 or more, one again has to apply the method recursively. Such an application is shown through an example with equation of degree 7.

To demonstrate how the proposed method works, it is considered on its own and no recourse is made to any of the known methods for the determination of the number of positive and negative roots of polynomial equations or to any techniques for the isolation of their roots, except for comparative purposes only.

This paper is structured as follows. The two variations of the method are presented in detail in Section 2. Section 3 provides examples of the second variation of the method and, on its basis, the full classification of the roots of the general cubic equation in terms of the equation coefficients is presented. Further examples of the application of this variation of the method are also given in Section 3. These include an equation of degree 5 and an equation of degree 7 and for the latter the second variation of the method is applied recursively. In Section 4, the first variation of the method is used recursively for an equation of degree 9. As polynomial equations very often arise in the study of various phenomena, the coefficients of these equations may be functions of the parameters of the model used for the analysis. The presence of even a single parameter as a coefficient in an equation of degree 3 or 4 makes the equation practically unsolvable, despite having explicit cubic and quartic formulae at hand. It is the location of the roots of such equations that can be the only source of beneficial information. In this respect, Section 5 shows the application of the first variation of the method to a quartic equation with coefficients which are intricate functions of the parameters of a model with which the wave–current interactions in the physically realistic model of azimuthal two-dimensional non-viscous flows with piecewise constant vorticity in a two-layer fluid with a flat bed and a free surface are studied. A two-tier analysis of this equation – analysis based on solving cubic equations and analysis based on solving quadratic equations only – is presented. Section 6 shows how the second variation of the method can be modified. Sections 7 and 8 provide step-by-step algorithms for the recursive application of each version of the method.

2 The method

Consider the equation

(1) a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 = 0

and write the corresponding polynomial p n ( x ) as

(2) p n ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 = f n ( x ) − g k − 1 ( x ) ,

where 0 < k < n and:

(3) f n ( x ) = x k a n x n − k + a n − 1 x n − k − 1 + ⋯ + a k + 1 x + a k ≡ x k F n − k ( x ) ,

(4) g k − 1 ( x ) = − a k − 1 x k − 1 − a k − 2 x k − 2 − ⋯ − a 1 x − a 0 .

The roots of the equation p n ( x ) = f n ( x ) − g k − 1 ( x ) = 0 can be found as the abscissas of the intersection points of the graphs of the polynomials f n ( x ) and g k − 1 ( x ) . The polynomial f n ( x ) has a zero f 0 of order k at the origin and if a k ≠ 0 and k is odd, it has a saddle there, while if a k ≠ 0 and k is even, f n ( x ) has a minimum or a maximum at the origin. The remaining roots of f n ( x ) = 0 are those of F n − k ( x ) = a n x n − k + a n − 1 x n − k − 1 + ⋯ + a k + 1 x + a k = 0 . The root f 0 = 0 , together with the real roots f i and g i of the lower-degree equations F n − k ( x ) = 0 and g k − 1 ( x ) = 0 , respectively, divides the abscissa into sub-intervals. The roots of the equation p n ( x ) = 0 can exist only in those sub-intervals where f n ( x ) and g k − 1 ( x ) have the same sign and this also allows the counting of the number of positive and negative roots of the given equation p n ( x ) = 0 . When counting, one should keep in mind that the function f n ( x ) can have up to n − 1 extremal points with the origin being an extremal point of order k − 1 , while the function g k − 1 ( x ) can have up to k − 2 extremal points. This places an upper limit on the number of sign changes of the first derivatives of f n ( x ) and g k − 1 ( x ) , that is, an upper limit on the “turns” which the polynomials f n ( x ) and g k − 1 ( x ) can do, and this, in turn, puts an upper limit on the count of the possible intersection points between f n ( x ) and g k − 1 ( x ) in the various sub-intervals. If there is an odd number of roots of f n ( x ) = 0 between two neighbouring roots of g k − 1 ( x ) = 0 (or, vice versa, if there is an odd number of roots of g k − 1 ( x ) = 0 between two neighbouring roots of f n ( x ) = 0 ), then there is an odd number of roots of the equation p n ( x ) = 0 between these two neighbouring roots of g k − 1 ( x ) = 0 (or between the two neighbouring roots of f n ( x ) = 0 ). But if there is an even number of roots of f n ( x ) = 0 between two neighbouring roots of g k − 1 ( x ) = 0 (or vice versa), then the number of roots of p n ( x ) = 0 between these two neighbouring roots of g k − 1 ( x ) = 0 (or between the two neighbouring roots of f n ( x ) = 0 ) is zero or some even number. In all cases, the end points of these sub-intervals serve as root bounds and thus the number of positive and negative roots can be found with significantly higher determinacy than the Descartes rule of signs provides.

All of the above is doable analytically for equations of degree up to and including 9 and is illustrated further with examples. In the case of p 9 ( x ) , one will only have to solve two equations of degree 4.

If one has a polynomial equation of degree 10 or higher, then the above procedure should be done recursively at the expense of reduced, but not at all exhausted, ability to determine root bounds and number of positive and negative roots.

To introduce a variation of the method, an assumption will be made: p n ( x ) is such that 0 is not among the roots of the corresponding polynomial equation p n ( x ) = 0 , that is a 0 ≠ 0 . As the determination of whether 0 is a root of an equation or not is absolutely straightforward, the case of a root being equal to zero will be of no interest for the analysis. In view of this, the coefficient a 0 will be set equal to 1. It suffices to say that should the equation p n ( x ) = 0 have zero as root of order m, then the remaining non-zero roots of the equation p n ( x ) = 0 can be found as the roots of the equation of degree n − m given by τ n − m ( x ) = p n ( x ) / x m = 0 .

One can consider an alternative split of the given polynomial p n ( x ) – it can be written as the difference of two polynomials, each of which passes through a fixed point in the ( x , y ) -plane:

(5) p n ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + 1 = q n ( x ) − r n − 1 ( x ) ,

where:

(6) q n ( x ) = a n x n + 1 ,

(7) r n − 1 ( x ) = − a n − 1 x n − 1 − a n − 2 x n − 2 − ⋯ − a 1 x .

The roots of the equation p n ( x ) = q n ( x ) − r n − 1 ( x ) = 0 are found as the abscissas of the intersection points of the graphs of the polynomials q n ( x ) and r n − 1 ( x ) . Regardless of its only coefficient a n , the polynomial q n ( x ) passes through point ( 0 , 1 ) – the reason behind the choice of a 0 = 1 – while the polynomial r n − 1 ( x ) passes through the origin (it has a zero there), regardless of the values of its coefficients a n − 1 , a n − 2 , … , a 1 .

The method will be illustrated first for this split.

Considering the given p n ( x ) , write α instead of the given coefficient a n and treat this α as undetermined. All other coefficients a j , j = 1 , 2 , … , n − 1 , are as they were given through the equation. Calculate next the discriminant Δ n ( α ) of the given polynomial p n ( x ) . If this discriminant is zero, then the equation p n ( x ) = 0 will have at least one repeated root. The equation Δ n ( α ) = 0 is an equation in α of degree n − 1 . Denote the n − 1 roots of the equation Δ n ( α ) = 0 by α 1 , α 2 … , α n − 1 . Then, for the real roots of Δ n ( α ) = 0 , each of the equations

(8) α j x n + a n − 1 x n − 1 + ⋯ + a 1 x + 1 = 0 , j ≤ n − 1 ,

will have a root β j of order at least 2. If, in each of the above equations, α j is perturbed slightly, so that Δ n ( α j ) becomes negative for that perturbed α j , then the double real root β j will become a pair of complex conjugate roots. If, instead, the perturbation of α j results in Δ n ( α j ) becoming positive, then the real double root β j will bifurcate into two different real roots – one on each side of β j .

It should be noted that if α j are all complex, then the equation p n ( x ) = 0 cannot have a repeated root, namely, the equation p n ( x ) = 0 has no real roots if it is of order 2 m or has just one real root if it is of order 2 m + 1 .

Consider now q n ( x ) and r n − 1 ( x ) . If the equation p n ( x ) = q n ( x ) − r n − 1 ( x ) = 0 has a double root χ , then the curves q n ( x ) and r n − 1 ( x ) will be tangent to each other at χ , that is q n ( χ ) = r n − 1 ( χ ) , and, also, the tangents to the curves q n ( x ) and r n − 1 ( x ) will coincide at χ , namely, q n ′ ( χ ) = r n − 1 ′ ( χ ) . The latter allows one to find

(9) a 1 = − 2 a 2 χ − 3 a 3 χ 2 − ⋯ − n a n χ n − 1 .

Substituting into q n ( χ ) = r n − 1 ( χ ) yields:

(10) − ( n − 1 ) a n χ n − ( n − 2 ) a n − 1 χ n − 1 − ⋯ − 2 a 3 χ 3 − a 2 χ 2 + 1 = 0 .

Adding ( n − 1 ) times q n ( χ ) − r n − 1 ( χ ) = 0 to the above results in the following equation:

(11) a n − 1 χ n − 1 + 2 a n − 2 χ n − 2 + ⋯ + ( n − 1 ) a 1 χ + n = 0 .

The roots χ j of this equation are the same as the double roots of Eq. (8). Eq. (11) is another equation of order one less than that of the original equation.

For an equation of degree up to and including 5, comparison of the given coefficient a 5 to the real numbers α j from the obtained set α 1 , α 2 , α 3 , α 4 allows not only the determination of the exact number of positive and negative roots, but also finding their isolation intervals.

For equation of degree 6 or higher, this variant of the method should be used recursively.

3 Examples for the split (5)

The method will now be illustrated with examples (of increasing complexity) of polynomial equations of different orders. In the trivial case of p 1 ( x ) = 0 , that is a x + 1 = 0 , the only root is determined by the intersection of the straight line y = a x + 1 with the abscissa y = 0 . The root x = − 1 / a always exists. It is positive when a < 0 and negative when a > 0 . The case of quadratic equation is also very simple – see Figure 1 for the full classification.

$Figure 1 The quadratic equation a x 2 + 1 = − b x a{x}^{2}+1=-bx . (a) The case of a > 0 a\gt 0 and b ≤ 0 b\le 0 . When a > 0 a\gt 0 and b ≥ 0 b\ge 0 , the situation is analogical to the one shown here as there is axial symmetry (reflection) with respect to the ordinate (one replaces b with − b -\hspace{-0.25em}b ). The points x 1 , 2 {x}_{1,2} are the roots of the quadratic equation. For fixed a, there is a value of b, say β \beta , such that at point x = χ x=\chi , the graphs of q 2 ( x ) {q}_{2}(x) and r 1 ( x ) {r}_{1}(x) are tangent. The tangents to the graphs at point x = χ x=\chi also coincide. Thus, q 2 ( x ) | ( x = χ , b = β ) = r 1 ( x ) | ( x = χ , b = β ) {q}_{2}(x){|}_{(x=\chi ,b=\beta )}={r}_{1}(x){|}_{(x=\chi ,b=\beta )} and ( d / d x ) q 2 ( x ) | ( x = χ , b = β ) = ( d / d x ) r 1 ( x ) | ( x = χ , b = β ) (\text{d}/\text{d}x){q}_{2}(x){|}_{(x=\chi ,b=\beta )}=(\text{d}/\text{d}x){r}_{1}(x){|}_{(x=\chi ,b=\beta )} . That is, one has the two simultaneous equations a χ 2 + 1 = − β χ a{\chi }^{2}+1=-\beta \chi and 2 a χ = − β 2a\chi =-\beta , from which it is easily determined that β = − 2 a \beta =-2\sqrt{a} and χ = − β / 2 a = 1 / a \chi =-\beta /2a=1/\sqrt{a} . The resulting root χ \chi of the quadratic equation is double. This corresponds to vanishing discriminant Δ 2 = b 2 − 4 a {\text{Δ}}_{2}={b}^{2}-4a . For − 2 a < b ≤ 0 -\hspace{-0.35em}2\sqrt{a}\lt b\le 0 (i.e. when the discriminant Δ 2 {\text{Δ}}_{2} is negative), there are no real roots of the quadratic equation (this includes the depressed equation a x 2 + 1 = 0 a{x}^{2}+1=0 for which b = 0 b=0 ). When b < − 2 a b\lt -2\sqrt{a} , the double root bifurcates into two: x 1 {x}_{1} and x 2 {x}_{2} so that they fall on either side of the double root χ \chi . For a > 0 a\gt 0 and b < 0 b\lt 0 , the roots, if they are real, are both positive. Thus, one of the roots of the quadratic equation a x 2 + b x + 1 = 0 a{x}^{2}+bx+1=0 is between the origin and 1 / a 1/\sqrt{a} . The other one is greater than 1 / a 1/\sqrt{a} . For a > 0 a\gt 0 and b < 0 b\lt 0 (the axially symmetric case), both roots, if they are real, are negative with one smaller than − 1 / a -\hspace{-0.25em}1/\sqrt{a} and the other between − 1 / a -\hspace{-0.25em}1/\sqrt{a} and the origin. (b) The case of a < 0 a\lt 0 with: b > 0 b\gt 0 , b = 0 b=0 (corresponding to the depressed equation a x 2 + 1 = 0 a{x}^{2}+1=0 ), and b < 0 b\lt 0 . The roots of the quadratic equation are real in each of the three sub-cases. One root is always positive, the other root is always negative. The roots of the suppressed equation are ± 1 / − a \pm \hspace{-0.25em}1/\sqrt{-a} (recall, a < 0 a\lt 0 now) Thus, the graph of the quadratic polynomial q 2 ( x ) = a x 2 + 1 {q}_{2}(x)=a{x}^{2}+1 always goes through point ( 0 , 1 ) (0,1) and also through points ± 1 / − a , 0 \left(\pm 1/\sqrt{-a},0\right) . When b < 0 b\lt 0 , the bigger root of the quadratic equation is between 0 and 1 / − a 1/\sqrt{-a} , while the smaller root is less than − 1 / − a -\hspace{-0.25em}1/\sqrt{-a} . When b > 0 b\gt 0 , the situation is symmetric: the smaller root is between − 1 / − a -\hspace{-0.25em}1/\sqrt{-a} and 0, while the bigger root is greater than 1/ − a \mathrm{1/}\sqrt{-a} . With a fixed and b → ∞ b\to \infty , the negative root tends to zero and the positive root tends to ∞ \infty . With a fixed and b → − ∞ b\to -\infty , the negative root tends to − ∞ -\hspace{-0.25em}\infty and the positive root tends to 0. Clearly, with a fixed and b → 0 ± b\to {0}^{\pm } , the roots tend to those of the suppressed equation. With b being a fixed positive number and a → 0 − a\to {0}^{-} , the positive root tends to + ∞ +\hspace{-0.25em}\infty and the negative root tends to − 1 / b -\hspace{-0.25em}1/b . With b being a fixed negative number and a → 0 − a\to {0}^{-} , the negative root tends to − ∞ -\hspace{-0.25em}\infty and the positive root tends to 1 / b 1/b . When b is a fixed number (positive, negative, or zero) and a → − ∞ a\to -\infty , the roots tend to 0 from either side, regardless of the sign of b.$

Figure 1

The quadratic equation a x 2 + 1 = − b x . (a) The case of a > 0 and b ≤ 0 . When a > 0 and b ≥ 0 , the situation is analogical to the one shown here as there is axial symmetry (reflection) with respect to the ordinate (one replaces b with − b ). The points x 1 , 2 are the roots of the quadratic equation. For fixed a, there is a value of b, say β , such that at point x = χ , the graphs of q 2 ( x ) and r 1 ( x ) are tangent. The tangents to the graphs at point x = χ also coincide. Thus, q 2 ( x ) | ( x = χ , b = β ) = r 1 ( x ) | ( x = χ , b = β ) and ( d / d x ) q 2 ( x ) | ( x = χ , b = β ) = ( d / d x ) r 1 ( x ) | ( x = χ , b = β ) . That is, one has the two simultaneous equations a χ 2 + 1 = − β χ and 2 a χ = − β , from which it is easily determined that β = − 2 a and χ = − β / 2 a = 1 / a . The resulting root χ of the quadratic equation is double. This corresponds to vanishing discriminant Δ 2 = b 2 − 4 a . For − 2 a < b ≤ 0 (i.e. when the discriminant Δ 2 is negative), there are no real roots of the quadratic equation (this includes the depressed equation a x 2 + 1 = 0 for which b = 0 ). When b < − 2 a , the double root bifurcates into two: x 1 and x 2 so that they fall on either side of the double root χ . For a > 0 and b < 0 , the roots, if they are real, are both positive. Thus, one of the roots of the quadratic equation a x 2 + b x + 1 = 0 is between the origin and 1 / a . The other one is greater than 1 / a . For a > 0 and b < 0 (the axially symmetric case), both roots, if they are real, are negative with one smaller than − 1 / a and the other between − 1 / a and the origin. (b) The case of a < 0 with: b > 0 , b = 0 (corresponding to the depressed equation a x 2 + 1 = 0 ), and b < 0 . The roots of the quadratic equation are real in each of the three sub-cases. One root is always positive, the other root is always negative. The roots of the suppressed equation are ± 1 / − a (recall, a < 0 now) Thus, the graph of the quadratic polynomial q 2 ( x ) = a x 2 + 1 always goes through point ( 0 , 1 ) and also through points ± 1 / − a , 0 . When b < 0 , the bigger root of the quadratic equation is between 0 and 1 / − a , while the smaller root is less than − 1 / − a . When b > 0 , the situation is symmetric: the smaller root is between − 1 / − a and 0, while the bigger root is greater than 1/ − a . With a fixed and b → ∞ , the negative root tends to zero and the positive root tends to ∞ . With a fixed and b → − ∞ , the negative root tends to − ∞ and the positive root tends to 0. Clearly, with a fixed and b → 0 ± , the roots tend to those of the suppressed equation. With b being a fixed positive number and a → 0 − , the positive root tends to + ∞ and the negative root tends to − 1 / b . With b being a fixed negative number and a → 0 − , the negative root tends to − ∞ and the positive root tends to 1 / b . When b is a fixed number (positive, negative, or zero) and a → − ∞ , the roots tend to 0 from either side, regardless of the sign of b.

3.1 Full classification of the roots of the cubic equation in terms of its coefficients

Despite the fact that there are formulae for the roots of the cubic (and quartic) equation, it is important to classify the roots in terms of the equation coefficients. This is especially valid for equations arising from various scientific models for which the equation coefficients may be functions of the model parameters. As the direct root determination may yield rather complicated expressions, knowledge of the root isolation intervals could be quite useful.

Consider a cubic equation which does not have a zero root. Without loss of generality, such equation can be written as p 3 ( x ) = a x 3 + b x 2 + c x + 1 = 0 or as either one of the following two splits: p 3 ( x ) = q 3 ( x ) − r 2 ( x ) = 0 with q 3 ( x ) = a x 3 + 1 and r 2 ( x ) = − b x 2 − c x , or p 3 ( x ) = ψ 3 ( x ) − φ 1 ( x ) = 0 with ψ 3 ( x ) = a x 3 + b x 2 and φ 1 ( x ) = − c x − 1 . That is:

(12) a x 3 + 1 = − b x 2 − c x ,

(13) a x 3 + b x 2 = − c x − 1 .

Provided that b ≠ 0 , the graph of the quadratic polynomial r 2 ( x ) = − b x 2 − c x passes through the origin and also through point − c / b from the abscissa. The graph of the cubic polynomial q 3 ( x ) = a x 3 + 1 passes through point ( 0 , 1 ) and also through point − 1 / a 3 from the abscissa. The cubic polynomial ψ 3 ( x ) = a x 3 + b x 2 has a double root at zero. If b is positive, the graph has a minimum at the origin, if b is negative, the graph has a maximum at the origin. The other zero of the cubic polynomial ψ 3 ( x ) = a x 3 + b x 2 is at − b / a . The graph of the polynomial φ 1 ( x ) = − c x − 1 is a straight line passing through point ( 0 , − 1 ) with slope − c .

3.1.1 The depressed cubic equation

First, the situation of the depressed cubic equation, i.e. equation with b = 0 , will be considered. In this case, the two splits become equivalent: the graphs of the pair in the split (12) are the same as the graphs of the pair in the split (13), but shifted vertically by one unit. To find for which α the equation α x 3 + c x + 1 = 0 would have a double root, consider the discriminant − α ( 4 c 3 + 27 α ) . This vanishes when α is either 0 or − 4 c 3 / 27 . The cubic equation α x 3 + c x + 1 = 0 with α = − 4 c 3 / 27 has root χ 0 = 3 / c and a double root χ = − 3 / ( 2 c ) .

One can then immediately determine the number of positive and negative roots of the original equation a x 3 + c x + 1 = 0 and also localize them as follows (Figure 2). If a and c are both positive, then the equation has one negative root x 0 located between the intersection point of the graph of a x 3 + 1 with the abscissa and the origin. Namely, − 1 / a 3 < x 0 < 0 . The other two roots are complex.

$Figure 2 The depressed cubic equation a x 3 + 1 = − c x a{x}^{3}+1=-cx . Equation α x 3 + 1 = − c x \alpha {x}^{3}+1=-cx with α = − 4 c 3 / 27 \alpha =-4{c}^{3}/27 has root χ 0 = 3 / c {\chi }_{0}=3/c and a double root χ = − 3 / ( 2 c ) \chi =-3/(2c) . Comparing the given a to α \alpha allows the immediate determination of the exact number of positive and negative roots of the equation and also their localization.$

Figure 2

The depressed cubic equation a x 3 + 1 = − c x . Equation α x 3 + 1 = − c x with α = − 4 c 3 / 27 has root χ 0 = 3 / c and a double root χ = − 3 / ( 2 c ) . Comparing the given a to α allows the immediate determination of the exact number of positive and negative roots of the equation and also their localization.

If a is positive and c is negative, then the equation has one negative root x 0 located to the left of the intersection point of the graph of a x 3 + 1 with the abscissa. Namely, x 0 < − 1 / a 3 . If, further, a > α , then the equation has two complex roots. If a = α , the equation has an additional double root given by χ = − 3 / 2 c > 0 . If, further, a < α , then the equation has, additionally, two positive roots: one between 0 and χ = − 3 / 2 c and the other greater than χ = − 3 / 2 c . One only needs to compare a, given through the equation, to α = − 4 c 3 / 27 for which the cubic discriminant vanishes.

If a is negative and c is positive, the situation is symmetric (with respect to the ordinate) to the situation of a positive and c negative. This time, the equation has one positive root x 0 located to the right of the intersection point of the graph of a x 3 + 1 with the abscissa. Namely, x 0 > − 1 / a 3 . If, further, a < α , then the equation has no other roots. If a = α , the equation has an additional double root given by χ = − 3 / 2 c < 0 . Finally, if a > α , then the equation has, additionally, two negative roots: one between 0 and χ = − 3 / 2 c and the other smaller than χ = − 3 / 2 c . Again, one only needs to compare a, given through the equation, to α = − 4 c 3 / 27 for which the cubic discriminant vanishes.

Finally, if both a and c are negative, then the equation has one positive root x 0 located between the origin and the intersection point of the graph of a x 3 + 1 with the abscissa, that is, 0 < x 0 < − 1 / a 3 .

As a numerical example of the above, consider the equation

(14) 3 x 3 − 4 x + 1 = 0 .

The auxiliary equation

(15) α x 3 − 4 x + 1 = 0

has discriminant − α ( 27 α − 256 ) . Clearly, when α = 256 / 27 , Eq. (15) will have a double root χ . To find χ , one needs to solve Eq. (11):

(16) 2 c χ + 3 = 0 ,

with c = − 4 . Thus, χ = 3 / 8 = 0.375 .

As the given value of a is 3 and as a = 3 < α = 256 / 27 ≈ 9.481 , the given Eq. (14) has two positive roots: x 2 which is between 0 and χ = 0.375 , and x 3 which is greater than χ = 0.375 . The equation also has a negative root x 1 to the left of the intersection point of the graph of 3 x 3 + 1 and the abscissa, that is x 1 < − 1 / 3 3 ≈ − 0.693 .

The roots of Eq. (14) are x 1 = − 1.264 , x 2 = 0.264 , x 3 = 1 .

3.1.2 The full cubic equation

If one considers next the cubic equation a x 3 + b x 2 + 1 = 0 , the situation will not turn out to be qualitatively different from the one of the “full” equation a x 3 + b x 2 + c x + 1 = 0 , where all coefficients are different from zero, and the latter is the equation to be considered next.

The discriminant of the cubic equation

(17) a x 3 + b x 2 + c x + 1 = 0

is given by

(18) Δ 3 = − 27 a 2 + 2 c ( 9 b − 2 c 2 ) a + b 2 ( c 2 − 4 b ) .

Setting Δ 3 = 0 and interpreting b and c as parameters, one gets a quadratic equation for a with roots

(19) α 1 , 2 = − 2 27 c 3 + 1 3 b c ± 2 27 ( c 2 − 3 b ) 3 2 .

These are real, i.e. the discriminant Δ 3 can be zero, only when c 2 > 3 b . Let α 1 denote the bigger root.

If, further, c 2 > 4 b , then the free term in the quadratic equation Δ 3 = 0 will be positive and, according to Viète’s formula, α 1 and α 2 will have different signs.

The two border cases are c 2 = 4 b , in which case the roots (19) are 0 and c 3 / 54 , and c 2 = 3 b , in which case equation Δ 3 = 0 has a double root c 3 / 27 .

For the case of a general equation of degree 3, Eq. (11) becomes:

(20) b x 2 + 2 c x + 3 = 0 .

The roots of this equation are

(21) χ 1 , 2 = 1 b − c ± c 2 − 3 b .

At point x = χ 1 , 2 , the curve α 1 , 2 x 3 + 1 is tangent to the curve − b x 2 − c x and the tangents to the graphs to each of the curves coincide at that point.

To determine the exact number of positive and negative roots and to also localize the roots, one has to compare the given a with the values of α 1 and α 2 .

Depending on the signs of the three parameters a, b, and c, there are eight cases to be analysed. Four will be considered in detail, and the analysis for each of the remaining four cases can be easily inferred afterwards.

3.1.2.1 The case of a > 0 , b > 0 , and c > 0

This is the most complicated case. There are three sub-cases.

3.1.2.1.1 The sub-case of − c / b < − 1 a 3

In this sub-case, the negative root of − b x 2 − c x is to the left of the point where a x 3 + 1 intersects the abscissa (Figure 3a).

$Figure 3 The cubic equation a x 3 + 1 = − b x 2 − c x a{x}^{3}+1=-b{x}^{2}-cx with a > 0 a\gt 0 , b > 0 b\gt 0 , and c > 0 c\gt 0 . (a) For the sub-case of − c / b < ( − a ) − 1 / 3 -\hspace{-0.25em}c/b\lt {(-a)}^{-1/3} , the smaller root − c / b -\hspace{-0.25em}c/b of − b x 2 − c x -\hspace{-0.25em}b{x}^{2}-cx is to the left of the intersection of a x 3 + 1 a{x}^{3}+1 with the abscissa. If, further, c 2 ≥ 4 b {c}^{2}\ge 4b , the quadratic equation Δ 3 = 0 {\text{Δ}}_{3}=0 will have roots α 1 > 0 {\alpha }_{1}\gt 0 and α 2 ≤ 0 {\alpha }_{2}\le 0 (the latter bears no relevance as a is considered positive in this case). Equation α 1 x 3 + 1 = − b x 2 − c x {\alpha }_{1}{x}^{3}+1=-b{x}^{2}-cx has a double negative root χ = ( c / b ) − 1 − 1 − 3 b / c 2 < − c / b \chi =(c/b)\left(-1-\sqrt{1-3b/{c}^{2}}\right)\lt -c/b and another negative root χ 0 = − 2 χ − b / α 1 {\chi }_{0}=-2\chi -b/{\alpha }_{1} . Then, given the restriction − c / b < − a − 1 / 3 -\hspace{-0.25em}c/b\lt -{a}^{-1/3} of this sub-case, i.e. a > ( b / c ) 3 a\gt {(b/c)}^{3} , one can have either: ( b / c ) 3 < a < α 1 {(b/c)}^{3}\lt a\lt {\alpha }_{1} which leads to the cubic equation having two negative roots x 1 {x}_{1} and x 2 {x}_{2} , such that x 1 < χ {x}_{1}\lt \chi and χ < x 2 < − c / b \chi \lt {x}_{2}\lt -c/b , together with another negative root x 3 {x}_{3} between the origin and the bigger root σ = − c / ( 2 b ) 1 − 1 − 4 b / c 2 < 0 \sigma =\left(-c/(2b)\right)\left(1-\sqrt{1-4b/{c}^{2}}\right)\lt 0 of the equation − b x 2 − c x = 1 -b{x}^{2}-cx=1 (the intersection point of a x 3 + 1 a{x}^{3}+1 with a = 0 a=0 and − b x 2 − c x -\hspace{-0.25em}b{x}^{2}-cx closer to the origin), i.e. σ < x 3 < 0 \sigma \lt {x}_{3}\lt 0 ; or one can have a = α 1 a={\alpha }_{1} in which case the cubic equation will have a negative root between σ \sigma and 0 and a double root at χ \chi ; or one can have a > α 1 a\gt {\alpha }_{1} , in which case the cubic equation will have a negative root between σ \sigma and 0 and two complex roots. If, instead of c 2 ≥ 4 b {c}^{2}\ge 4b , one has c 2 < 3 b {c}^{2}\lt 3b (see Figure 3b for the case of 3 b < c 2 < 4 b 3b\lt {c}^{2}\lt 4b ), then the cubic discriminant Δ 3 {\text{Δ}}_{3} will be negative and the cubic equation will have a negative root between − a − 1 / 3 -\hspace{-0.25em}{a}^{-1/3} and 0 and two complex roots. If c 2 = 3 b {c}^{2}=3b , then the double root χ \chi is at − c / b = − 3 / c -\hspace{-0.25em}c/b=-3/c and α 1 = α 2 = c 3 / 27 {\alpha }_{1}={\alpha }_{2}={c}^{3}/27 . (b) If 3 b < c 2 < 4 b 3b\lt {c}^{2}\lt 4b and with no further restrictions except a > 0 , b > 0 a\gt 0,b\gt 0 , and c > 0 c\gt 0 , the roots α 1 , 2 {\alpha }_{1,2} of the quadratic equation Δ 3 = 0 {\text{Δ}}_{3}=0 will have the same sign. In this case, the coefficient 2 c ( 9 b − 2 c 2 ) 2c(9b-2{c}^{2}) in the term linear in a will be positive and thus both roots α 1 {\alpha }_{1} and α 2 {\alpha }_{2} will be positive (with α 1 > α 2 {\alpha }_{1}\gt {\alpha }_{2} ). The curve α 1 x 3 + 1 {\alpha }_{1}{x}^{3}+1 will be tangent to the curve − b x 2 − c x -\hspace{-0.25em}b{x}^{2}-cx at point χ 1 < − c / b {\chi }_{1}\lt -c/b , while curve α 2 x 3 + 1 {\alpha }_{2}{x}^{3}+1 will be tangent to the curve − b x 2 − c x -\hspace{-0.25em}b{x}^{2}-cx at point − c / b < χ 2 < − c / ( 2 b ) -\hspace{-0.25em}c/b\lt {\chi }_{2}\lt -c/(2b) . The points χ 1 , 2 {\chi }_{1,2} are the roots of the equation b x 2 + 2 c x + 3 = 0 b{x}^{2}+2cx+3=0 , that is χ 1 , 2 = ( c / b ) − 1 ± 1 − 3 b / c 2 {\chi }_{1,2}=(c/b)\left(-1\pm \sqrt{1-3b/{c}^{2}}\right) . Then, the roots of the cubic equation a x 3 + b x 2 + c x + 1 = 0 a{x}^{3}+b{x}^{2}+cx+1=0 will be as follows. If the given a is such that a > α 1 a\gt {\alpha }_{1} , the cubic equation will have a negative root between − a − 1 / 3 -\hspace{-0.25em}{a}^{-1/3} and 0 and two complex roots. If a = α 1 a={\alpha }_{1} , there will be a negative root between − α 1 − 1 / 3 -\hspace{-0.25em}{\alpha }_{1}^{-1/3} and 0 and a negative double root at χ 1 {\chi }_{1} . If a is such that α 2 < a < α 1 {\alpha }_{2}\lt a\lt {\alpha }_{1} , the cubic equation will have three negative roots: x 1 < χ 1 , χ 1 < x 2 < χ 2 {x}_{1}\lt {\chi }_{1},{\chi }_{1}\lt {x}_{2}\lt {\chi }_{2} , and χ 2 < x 3 < 0 {\chi }_{2}\lt {x}_{3}\lt 0 . If a = α 2 a={\alpha }_{2} , there will be a negative root smaller than − α 2 − 1 / 3 -\hspace{-0.25em}{\alpha }_{2}^{-1/3} and a negative double root at χ 2 {\chi }_{2} . Finally, if a < α 2 a\lt {\alpha }_{2} the cubic equation will have a negative root smaller than − a − 1 / 3 -\hspace{-0.25em}{a}^{-1/3} and two complex roots. If c 2 = 3 b {c}^{2}=3b , then α 1 {\alpha }_{1} and α 2 {\alpha }_{2} coalesce to c 3 / 27 {c}^{3}/27 while χ 1 {\chi }_{1} and χ 2 {\chi }_{2} coalesce to − c / b = − 3 / c -\hspace{-0.25em}c/b=-3/c . If c 2 < 3 b {c}^{2}\lt 3b , then χ 1,2 {\chi }_{\mathrm{1,2}} are not among the reals. If c 2 = 4 b {c}^{2}=4b , then α 2 {\alpha }_{2} becomes 0 and χ 2 = − c / ( 2 b ) {\chi }_{2}=-c/(2b) – the point of the maximum of − b x 2 − c x -\hspace{-0.25em}b{x}^{2}-cx . When c 2 > 4 b {c}^{2}\gt 4b , one has χ 2 > − c / ( 2 b ) {\chi }_{2}\gt -c/(2b) and α 2 < 0 {\alpha }_{2}\lt 0 .$

Figure 3

The cubic equation a x 3 + 1 = − b x 2 − c x with a > 0 , b > 0 , and c > 0 . (a) For the sub-case of − c / b < ( − a ) − 1 / 3 , the smaller root − c / b of − b x 2 − c x is to the left of the intersection of a x 3 + 1 with the abscissa. If, further, c 2 ≥ 4 b , the quadratic equation Δ 3 = 0 will have roots α 1 > 0 and α 2 ≤ 0 (the latter bears no relevance as a is considered positive in this case). Equation α 1 x 3 + 1 = − b x 2 − c x has a double negative root χ = ( c / b ) − 1 − 1 − 3 b / c 2 < − c / b and another negative root χ 0 = − 2 χ − b / α 1 . Then, given the restriction − c / b < − a − 1 / 3 of this sub-case, i.e. a > ( b / c ) 3 , one can have either: ( b / c ) 3 < a < α 1 which leads to the cubic equation having two negative roots x 1 and x 2 , such that x 1 < χ and χ < x 2 < − c / b , together with another negative root x 3 between the origin and the bigger root σ = − c / ( 2 b ) 1 − 1 − 4 b / c 2 < 0 of the equation − b x 2 − c x = 1 (the intersection point of a x 3 + 1 with a = 0 and − b x 2 − c x closer to the origin), i.e. σ < x 3 < 0 ; or one can have a = α 1 in which case the cubic equation will have a negative root between σ and 0 and a double root at χ ; or one can have a > α 1 , in which case the cubic equation will have a negative root between σ and 0 and two complex roots. If, instead of c 2 ≥ 4 b , one has c 2 < 3 b (see Figure 3b for the case of 3 b < c 2 < 4 b ), then the cubic discriminant Δ 3 will be negative and the cubic equation will have a negative root between − a − 1 / 3 and 0 and two complex roots. If c 2 = 3 b , then the double root χ is at − c / b = − 3 / c and α 1 = α 2 = c 3 / 27 . (b) If 3 b < c 2 < 4 b and with no further restrictions except a > 0 , b > 0 , and c > 0 , the roots α 1 , 2 of the quadratic equation Δ 3 = 0 will have the same sign. In this case, the coefficient 2 c ( 9 b − 2 c 2 ) in the term linear in a will be positive and thus both roots α 1 and α 2 will be positive (with α 1 > α 2 ). The curve α 1 x 3 + 1 will be tangent to the curve − b x 2 − c x at point χ 1 < − c / b , while curve α 2 x 3 + 1 will be tangent to the curve − b x 2 − c x at point − c / b < χ 2 < − c / ( 2 b ) . The points χ 1 , 2 are the roots of the equation b x 2 + 2 c x + 3 = 0 , that is χ 1 , 2 = ( c / b ) − 1 ± 1 − 3 b / c 2 . Then, the roots of the cubic equation a x 3 + b x 2 + c x + 1 = 0 will be as follows. If the given a is such that a > α 1 , the cubic equation will have a negative root between − a − 1 / 3 and 0 and two complex roots. If a = α 1 , there will be a negative root between − α 1 − 1 / 3 and 0 and a negative double root at χ 1 . If a is such that α 2 < a < α 1 , the cubic equation will have three negative roots: x 1 < χ 1 , χ 1 < x 2 < χ 2 , and χ 2 < x 3 < 0 . If a = α 2 , there will be a negative root smaller than − α 2 − 1 / 3 and a negative double root at χ 2 . Finally, if a < α 2 the cubic equation will have a negative root smaller than − a − 1 / 3 and two complex roots. If c 2 = 3 b , then α 1 and α 2 coalesce to c 3 / 27 while χ 1 and χ 2 coalesce to − c / b = − 3 / c . If c 2 < 3 b , then χ 1,2 are not among the reals. If c 2 = 4 b , then α 2 becomes 0 and χ 2 = − c / ( 2 b ) – the point of the maximum of − b x 2 − c x . When c 2 > 4 b , one has χ 2 > − c / ( 2 b ) and α 2 < 0 .

If one further has c 2 ≥ 4 b , then α 1 > 0 and α 2 ≤ 0 . The non-positive root α 2 is not relevant to the analysis as a, given through the equation, is positive in this case (the curve α 2 x 3 + 1 with α 2 ≤ 0 is also tangent to the curve − b x 2 − c x , but the given equation a x 3 + 1 = − b x 2 − c x has a > 0 ). Eq. (11) has root χ 1 = ( c / b ) − 1 − 1 − 3 b / c 2 < − c / b < 0 and root χ 2 , which is associated with α 2 . Thus, the double root χ of equation α 2 x 3 + 1 = − b x 2 − c x will be given by χ 1 , while the other root (using Viète’s formula) will be χ 0 = − 2 χ − b / α 1 .

When c 2 = 4 b , the curve a x 3 + 1 with a = 0 is tangent to − b x 2 − c x at the maximum c 2 / ( 4 b ) of − b x 2 − c x which occurs at − c / ( 2 b ) . When c 2 > 4 b , the curve a x 3 + 1 with a = 0 will intersect the curve − b x 2 − c x between − c / ( 2 b ) and the origin at point σ which is the bigger root of the quadratic equation − b x 2 − c x = − 1 , that is − c / ( 2 b ) < σ = [ − c / ( 2 b ) ] 1 − 1 − 4 b / c 2 < 0 . Thus, for any a greater than 0, provided that − c / b < − 1 / a 3 , the curve a x 3 + 1 will intersect the curve − b x 2 − c x , for which c 2 ≥ 4 b , once between σ and 0.

As a sub-case with − c / b < − 1 / a 3 is being studied, one has a > ( b / c ) 3 . If ( b / c ) 3 < a < α 1 , the cubic equation will have two negative roots x 1 and x 2 , such that x 1 < χ and χ < x 2 < − c / b , where χ = ( c / b ) − 1 − 1 − 3 b / c 2 and − c / b is the smaller root of − b x 2 − c x = 0 , and another negative root x 3 between σ and the origin, i.e. σ < x 3 < 0 . If a → α 1 from below, then the roots x 1 and x 2 will tend to χ from either side until they coalesce at the double root χ when a = α 1 . When a > α 1 , there will be a negative root between σ and 0 and two complex roots.

Consider as numerical examples for equation with a > 0 , b > 0 , c > 0 , c 2 ≥ 4 b , and − c / b < − 1 / a 3 the following two equations.

(22) x 3 + 2 x 2 + 3 x + 1 = 0 .

The roots α 1 , 2 are ± 2 3 / 9 . The relevant one is α 1 = 2 3 / 9 ≈ 0.385 . The roots χ 1 , 2 are − ( 3 / 2 ) ± 3 / 2 . The one of interest is χ = − ( 3 / 2 ) ± 3 / 2 ≈ − 2.366 . For this equation, one has σ = [ − c / ( 2 b ) ] 1 − 1 − 4 b / c 2 = − 1 / 2 . As a = 1 > α 1 ≈ 0.385 , the given equation has one negative root between σ = − 0.5 and 0 and two complex roots. Indeed, the roots are − 0.430 and − 0.785 ± 1.307 i approximately.

For the equation

(23) 1 3 x 3 + 2 x 2 + 3 x + 1 = 0 ,

one again has α 1 = 2 3 / 9 ≈ 0.385 , χ = − ( 3 / 2 ) ± 3 / 2 ≈ − 2.366 , and σ = − 1 / 2 . This time a = 1 / 3 < α 1 ≈ 0.385 . Therefore, the given equation has three negative roots: one smaller than χ ≈ − 2.366 , another one between χ ≈ − 2.366 and − c / b = − 1.5 , and the third one between σ = − 0.5 and the origin. Indeed, the roots are − 3.879 , − 1.653 , and − 0.468 approximately.

If one has 3 b ≤ c 2 < 4 b instead of c 2 ≥ 4 b , the roots α 1 , 2 are both positive (with α 1 = α 2 when c 2 = 3 b , otherwise α 1 > α 2 ). The curve α 1 x 3 + 1 will be tangent to the curve − b x 2 − c x at point χ 1 ≤ − c / b , while curve α 2 x 3 + 1 will be tangent to the curve − b x 2 − c x at point − ( c / b ) ≤ χ 2 < − c / ( 2 b ) . The points χ 1 , 2 are the roots of the equation b x 2 + 2 c x + 3 = 0 , that is χ 1 , 2 = ( c / b ) − 1 ± 1 − 3 b / c 2 (Figure 3b). If c 2 = 3 b , then α 1 and α 2 coalesce to c 3 / 27 , while χ 1 and χ 2 coalesce to − c / b = − 3 / c . Then, the roots of the cubic equation a x 3 + b x 2 + c x + 1 = 0 will be as follows. If the given a is such that a > α 1 , the cubic equation will have a negative root between − 1 / a 3 and 0 and two complex roots. If a = α 1 , there will be a negative root between − 1 / α 1 3 and 0 and a negative double root at χ 1 . If α 2 < a < α 1 , the cubic equation will have three negative roots x 1 , 2 , 3 such that x 1 < χ 1 , χ 1 < x 2 < χ 2 , and χ 2 < x 3 < 0 . If a = α 2 , there will be a negative root smaller than − 1 / α 2 3 and a negative double root at χ 2 . Finally, if a < α 2 the cubic equation will have a negative root smaller than − 1 / a 3 and two complex roots.

As the current sub-case has the restriction − c / b < − 1 / a 3 , one can only have a > ( b / c ) 3 . Thus, when ( b / c ) 3 < a < α 1 , the cubic equation will have two negative roots x 1 and x 2 , such that x 1 < χ 1 and χ 1 < x 2 < − c / b , and another negative root x 3 between the intersection point of a x 3 + 1 with the abscissa and the origin: − 1 / a 3 < x 3 < 0 . When a > α 1 , the cubic equation will have one negative root between − 1 / a 3 and the origin together with two complex roots.

As numerical examples for the equation with a > 0 , b > 0 , c > 0 , 3 b ≤ c 2 < 4 b , and − c / b < − 1 / a 3 consider the following two equations.

(24) x 3 + 5 3 x 2 + 23 10 x + 1 = 0 .

The roots α 1,2 are 5083 ± 29 29 / 13 , 500 . That is, α 1 ≈ 0.388 and α 2 ≈ 0.365 . The roots χ 1 , 2 are − 69 ± 3 29 / 50 , namely, χ 1 ≈ − 1.703 and χ 2 ≈ − 1.057 . As in the given equation a = 1 , which is greater than the bigger root α 1 ≈ 0.388 , the equation has a negative root between − 1 / a 3 and the origin, i.e. between − 1 and 0, and two complex roots. Indeed, the roots of the equation are − 0.603 and − 0.532 ± 1.173 i approximately.

Next, for the equation

(25) 385 1 , 000 x 3 + 5 3 x 2 + 23 10 x + 1 = 0 ,

the roots α 1 , 2 are the same: α 1 ≈ 0.388 and α 2 ≈ 0.365 . The roots χ 1 , 2 are also the same: χ 1 ≈ − 1.703 and χ 2 ≈ − 1.057 . The smaller root of − b x 2 − c x = 0 is − c / b = − 1.38 – between χ 1 and χ 2 , as expected. Also in this case: ( b / c ) 3 ≈ 0.381 . As the given a is 0.385, one has α 2 < ( b / c ) 3 < a < α 1 . Therefore, the roots of this cubic equation must be all negative and such that: one is smaller than χ 1 ≈ − 1.703 ; another one is between χ 1 ≈ − 1.703 and − c / b = − 1.38 ; and the third one is between − 1 / a 3 ≈ − 1.375 and 0. Indeed, the roots are − 1.938 , − 1.494 , and − 0.897 approximately.

Finally, when c 2 < 3 b in the sub-case of − c / b < − 1 / a 3 , then the cubic equation will have one negative root between − 1 / a 3 and 0 and two complex roots. This is illustrated by

(26) 2 x 3 + x 2 + x + 1 = 0 .

The model predicts a negative root between − 1 / a 3 ≈ − 0.794 and 0 and two complex roots. Indeed, one has x 1 = − 0.739 and x 2 , 3 = 0.119 ± 0.814 i .

3.1.2.1.2 The sub-case of − c / b > − 1 a 3

In this sub-case, the negative root of − b x 2 − c x is between the point where a x 3 + 1 intersects the abscissa and the origin.

If in this case one also has c 2 > 4 b , then the curve a x 3 + 1 with a = 0 will intersect the curve − b x 2 − c x between − c / b and the origin at two points: the roots σ 1 , 2 of the quadratic equation − b x 2 − c x = 1 , i.e. σ 1 , 2 = [ − c / ( 2 b ) ] 1 ± 1 − 4 b / c 2 < 0 (with σ 2 < σ 1 ). These are on either side of − c / ( 2 b ) . Then for any a greater than 0, given that − c / b > − 1 / a 3 , the curve a x 3 + 1 will intersect the curve − b x 2 − c x with c 2 ≥ 4 b , once between − c / b and σ 2 and one more time between σ 1 and the origin. There will be a third intersection to the left of − 1 / a 3 . Therefore, the cubic equation will have three negative roots, the biggest of which will be between σ 1 and the origin, the middle one will be between − c / b and σ 2 , and the smallest one will be to the left of − 1 / a 3 . A double root χ cannot exist when c 2 ≥ 4 b .

Equation with a > 0 , b > 0 , c > 0 , c 2 ≥ 4 b , and − c / b > − 1 / a 3 can be illustrated with the following numerical example:

(27) 1 30 x 3 + x 2 + 3 x + 1 = 0 .

The roots are x 1 ≈ − 26.667 , x 2 ≈ − 2.952 , and x 3 ≈ − 0.381 and within their predicted bounds: the biggest one is between σ 1 ≈ − 0.382 and the origin, the middle one is between − c / b = − 3 and σ 2 ≈ − 2.618 , and the third one is less than − 1 / a 3 ≈ − 3.107 .

Next, if one has 3 b ≤ c 2 < 4 b instead of c 2 ≥ 4 b , the situation in Figure 3c applies. In view of the restriction − c / b > − 1 / a 3 , one can have a < α 2 , a = α 2 , or α 2 < a < ( b / c ) 3 . In the first case, the cubic equation will have a negative root smaller than − 1 / a 3 and two complex roots. In the second case, the cubic equation will have a negative root smaller than − 1 / a 3 and a double negative root at χ 2 . In the third case, the cubic equation will have three negative roots: x 1 < χ 1 , χ 1 < x 2 < χ 2 , and χ 2 < x 3 < 0 . (As in the previous sub-case, if c 2 = 3 b , then α 1 and α 2 coalesce to c 3 / 27 , while χ 1 and χ 2 coalesce to − c / b = − 3 / c .)

Three numerical examples for equation with a > 0 , b > 0 , c > 0 , 3 b ≤ c 2 < 4 b , and − c / b > − 1 / a 3 are given. The first one is as follows:

(28) 2 5 x 3 + 2 x 2 + 5 2 x + 1 = 0 .

For this equation, the roots α 1 , 2 are 14/27 and 1/2, i.e. α 1 ≈ 0.519 , α 2 = 0.5 . The respective loci of the double roots are χ 1 = − 3 / 2 and χ 2 = − 1 . In this case, the given a = 2 / 5 = 0.4 is a number smaller than α 2 = 0.5 . Therefore, the equation has a negative root smaller than − 1 / a 3 ≈ − 1.357 and two complex roots. The roots of the equation are x 1 = − 3.362 and x 2 , 3 = − 0.819 ± 0.270 i .

Next, for the equation

(29) 1 2 x 3 + 2 x 2 + 5 2 x + 1 = 0 ,

one has α 1 ≈ 0.519 and α 2 = 0.5 with χ 1 = − 3 / 2 and χ 2 = − 1 . The given a = 1 / 2 is exactly equal to α 2 = 0.5 . Thus, there must be a negative root smaller than − 1 / a 3 ≈ − 1.256 and a double root at χ 2 = − 1 . The roots of this equation are − 2 , − 1 , and − 1 – exactly in the predicted bounds.

In the third example, the equation

(30) 51 100 x 3 + 2 x 2 + 5 2 x + 1 = 0

also has α 1 ≈ 0.519 and α 2 = 0.5 with χ 1 = − 3 / 2 and χ 2 = − 1 . In this case, the given a = 51 / 100 is a number between α 2 = 0.5 and ( b / c ) 3 = 0.512 . The model predicts that the equation will have three negative roots such that the smallest one is smaller than χ 1 = − 1.5 , the middle one is between χ 1 = − 1.5 and χ 2 = − 1 , and the third one is between χ 2 = − 1 and 0. Indeed, the roots are − 1.821 , − 1.213 , and − 0.888 approximately.

Finally, when c 2 < 3 b in the sub-case of − c / b > − 1 / a 3 , then there is one negative root smaller than − 1 / a 3 and two complex roots. This is illustrated by the equation:

(31) 1 2 x 3 + x 2 + x + 1 = 0 .

The negative root is x 1 ≈ − 1.544 < − 1 / a 3 ≈ − 1.260 and the other two roots are complex: x 2 , 3 ≈ − 0.228 ± 0.115 i – as predicted.

3.1.2.1.3 The sub-case of − c / b = − 1 a 3

When a = b 3 / c 3 , the cubic discriminant is Δ 3 = ( 9 b 2 / c 6 ) ( c 2 + b ) ( c 2 − 3 b ) . It is negative when c 2 < 3 b and in this case the cubic equation will have a negative root given by x = − c / b = − 1 / a 3 together with two complex roots. If the discriminant Δ 3 is positive (i.e. c 2 > 3 b ), the cubic equation will have three negative roots given by: x 1 < − c / b = − 1 / a 3 , x 2 = − c / b = − 1 / a 3 , and x 3 > − c / b = − 1 / a 3 . Finally, when the discriminant Δ 3 is zero, one will have b = c 2 / 3 . Given that a = b 3 / c 3 , one gets that a = c 3 / 27 . Thus, the cubic equation c 3 x 3 / 27 + c 2 x 2 / 3 + c x + 1 = 0 will have a triple negative root at − 3 / c .

The following three examples illustrate all possibilities for this sub-case.

First, the equation

(32) x 3 + x 2 + x + 1 = 0 ,

which is in the category c 2 < 3 b with a = ( b / c ) 3 , should have a negative root given by x = − c / b = − 1 together with two complex roots. This is the case indeed, as the roots are x 1 = 1 , x 2 , 3 = ± i .

Next, the equation

(33) 8 27 x 3 + 2 x 2 + 3 x + 1 = 0

is with c 2 > 3 b and, since a = ( b / c ) 3 , the roots are all negative: one is smaller than − c / b = − 1.5 , another one is equal to − c / b = − 1.5 , and the third one is between − c / b = − 1.5 and 0. The roots of the equation are x 1 ≈ − 4.780 , x 2 = − 1.5 , and x 3 = − 0.471 , which confirms the prediction of the model.

Finally, for the equation

(34) 8 27 x 3 + 4 3 x 2 + 2 x + 1 = 0

one has c = 3 b 2 and, given that a = ( b / c ) 3 , there must be a triple root given by − 3 / c = − 3 / 2 . This is so indeed.

3.1.2.2 The case of a > 0 , b > 0 , and c < 0

This case is illustrated in Figure 4a. Consider first the sub-case of c 2 > 4 b . The roots α 1 , 2 in (19) have different signs ( α 1 > 0 ). There can be only one curve α 1 x 3 + 1 , with positive α 1 , tangent to the curve − b x 2 − c x . This occurs at point χ = ( c / b ) − 1 − 1 − 3 b / c 2 < − c / b < 0 , which is the smaller root of Eq. (11). The other root, ( c / b ) − 1 + 1 − 3 b / c 2 < − c / b > 0 , is associated with the irrelevant α 2 < 0 .

$Figure 4 The cubic equation a x 3 + 1 = − b x 2 − c x a{x}^{3}+1=-b{x}^{2}-cx : the cases of a > 0 , b > 0 , c < 0 a\gt 0,b\gt 0,c\lt 0 (sub-figure (a)); a > 0 , b < 0 , c > 0 a\gt 0,b\lt 0,c\gt 0 (sub-figure (b)); and a > 0 , b < 0 , c < 0 a\gt 0,b\lt 0,c\lt 0 (sub-figure (c)).$

Figure 4

The cubic equation a x 3 + 1 = − b x 2 − c x : the cases of a > 0 , b > 0 , c < 0 (sub-figure (a)); a > 0 , b < 0 , c > 0 (sub-figure (b)); and a > 0 , b < 0 , c < 0 (sub-figure (c)).

When a is smaller than α 1 , the biggest root is positive and is between χ and σ 1 = [ − c / ( 2 b ) ] 1 + 1 − 4 b / c 2 , which is the bigger root of − b x 2 − c x − 1 = 0 . The middle root is also positive and is between σ 2 = [ − c / ( 2 b ) ] 1 − 1 − 4 b / c 2 (which is the smaller root of − b x 2 − c x − 1 = 0 ) and χ . The third root is negative and is smaller than − 1 / a 3 .

If a = α 1 , there is a positive double root χ and a negative root smaller than − 1 / a 3 .

Finally, if a > α 1 , then the cubic discriminant Δ 3 is negative and the equation has a negative root smaller than − 1 / a 3 together with two complex roots.

When c 2 = 4 b , the bigger root in (19) is α 1 = 0 . The other one is α 2 = c 3 / 54 < 0 . Thus, the curve a x 3 + 1 with a = α 1 = 0 is tangent to the curve − b x 2 − c x at point − c / ( 2 b ) , where the maximum of − b x 2 − c x occurs (the maximum in this case is c 2 / 4 b = 1 ). As a cannot be zero (one has to have an equation of degree 3), the cubic equation has a negative root smaller than − 1 / a 3 and two complex roots.

If 3 b ≤ c 2 < 4 b , the roots α 1 , 2 will have the same sign. The coefficient in the term linear in a is 2 c ( 9 b − 2 c 2 ) and in the case of 3 b ≤ c 2 < 4 b , given that b is positive, its sign will depend on the sign of c only. Thus, for negative c the roots α 1 and α 2 will be negative and thus irrelevant. No curve a x 3 + 1 with a > 0 can intersect in the first quadrant the curve − b x 2 − c x with c 2 ≤ 4 b . The cubic discriminant Δ 3 is non-negative. The roots of the cubic equation are as in the latter case: two complex and one negative and smaller than − 1 / a 3 .

When c 2 < 3 b , the discriminant Δ 3 is negative and no curve a x 3 + 1 with any a could be tangent to the curve − b x 2 − c x . The roots of the cubic equation are, again, two complex and one negative and smaller than − 1 / a 3 .

This case will be illustrated with the following four examples.

Consider first the equation

(35) x 3 + x 2 − 5 x + 1 = 0 .

The roots (19) are α 1 = 205 / 27 + ( 44 / 27 ) 22 ≈ 15.236 and α 2 = 205 / 27 − ( 44 / 27 ) 22 ≈ − 0.051 (irrelevant). The relevant root χ = χ 1 is 5 − 22 ≈ 0.310 . The roots of the equation − b x 2 − c x − 1 = 0 are σ 1 = 5 / 2 + ( 1 / 2 ) 21 ≈ 4.791 and σ 2 = 5 / 2 − ( 1 / 2 ) 21 ≈ 0.209 . As the given a is equal to 1 and smaller than α 1 ≈ 15.236 , and as the given b and c are such that c 2 > 4 b , then the roots of the equation are as follows: a positive root x 1 between χ ≈ 0.310 and σ 1 ≈ 4.791 , another positive root ( x 2 ) between σ 2 ≈ 0.209 and χ ≈ 0.310 , and a negative root x 3 smaller than − 1 / a 3 = − 1 . The roots of the equation are x 1 ≈ 1.655 , x 2 ≈ 0.211 , and x 3 ≈ − 2.866 , which agrees with the prediction.

Next, for the equation

(36) 16 x 3 + x 2 − 5 x + 1 = 0 ,

the corresponding equation Δ 3 = 0 has the same roots α 1 = 205 / 27 + ( 44 / 27 ) 22 ≈ 15.236 and α 2 = 205 / 27 − ( 44 / 27 ) 22 ≈ − 0.051 (irrelevant). The relevant root χ = χ 1 is 5 − 22 ≈ 0.310 is the same. The roots of the equation − b x 2 − c x − 1 = 0 are also the same: σ 1 = 5 / 2 + ( 1 / 2 ) 21 ≈ 4.791 and σ 2 = 5 / 2 − ( 1 / 2 ) 21 ≈ 0.209 . As the given a is now equal to 16 and greater than α 1 ≈ 15.236 , and as the given b and c are still such that c 2 > 4 b , then the roots of the equation are as follows: a positive root x 1 between χ ≈ 0.310 and σ 1 ≈ 4.791 , another positive root ( x 2 ) between σ 2 ≈ 0.209 and χ ≈ 0.310 , and a negative root x 3 smaller than − 1 / a 3 ≈ − 0.397 . The roots of the equation are x 1 , 2 ≈ 0.303 ± 0.375 i and x 3 ≈ 0.669 , which also agrees with the prediction.

As a further example, for the equation

(37) x 3 + 2 x 2 − 13 5 x + 1 = 0 ,

the corresponding equation Δ 3 = 0 has roots α 1 = − 1 , 456 / 3 , 375 + ( 38 / 3 , 375 ) 19 ≈ − 0.382 and α 2 = − 1 , 456 / 3 , 375 − ( 38 / 3 , 375 ) 19 ≈ − 0.480 . Both are irrelevant, since the only curves a x 3 + 1 that can be tangent to − b x 2 − c x are those with a < 0 , while the considered a is positive. The equation should have two complex roots and a negative root smaller than − 1 / a 3 = − 1 . This is the case indeed: x 1 , 2 ≈ 0.492 ± 0.305 i and x 3 ≈ − 2.984 .

Finally, the equation

(38) 3 x 3 + 2 x 2 − x + 1 = 0

is with c 2 < 3 b . The discriminant Δ 3 is negative. Thus, there should be a root smaller than − 1 / a 3 ≈ − 0.693 and two complex roots. This is the case indeed: x 1 ≈ − 1.185 and x 1 , 2 ≈ 0.259 ± 0.463 i .

3.1.2.3 The case of a > 0 , b < 0 , and c > 0

Given that b is negative, real roots (19) always exist and they always are with opposite signs. The relevant one is α 1 > 0 . The corresponding root of (11) is χ = χ 1 = ( c / b ) − 1 + 1 − 3 b / c 2 > − c / b > 0 . The curve a x 3 + 1 with a = 0 intersects in the first quadrant the curve − b x 2 − c x with b < 0 and c > 0 at point σ 1 = [ − c / ( 2 b ) ] 1 + 1 − 4 b / c 2 , which is the bigger root of the equation − b x 2 − c x − 1 = 0 .

Then the roots of the cubic equation are as follows. There is always one negative root between − 1 / a 3 and the origin. If a > α 1 , then the other two roots are complex. If a = α 1 , then in addition to the negative root between − 1 / a 3 and 0, there is a positive double root at χ . If a < α 1 , the roots are one negative root between − 1 / a 3 and 0; one positive root between σ 1 and χ ; and another positive root greater than χ (Figure 4b).

The equation

(39) x 3 − x 2 + x + 1 = 0

illustrates the case of a > α 1 : one has a = 1 > α 1 = 5 / 27 ≈ 0.185 . The method predicts one negative root between − 1 / a 3 = − 1 and 0 and two complex roots. The roots are x 1 ≈ − 0.544 and x 2 , 3 ≈ 0.772 ± 1.115 i .

The equation

(40) 1 9 x 3 − x 2 + x + 1 = 0

is in the category of a = 1 / 9 < α 1 = 5 / 27 ≈ 0.185 . The locus of the corresponding double root is χ = χ 1 = 3 . The bigger root of − b x 2 − c x − 1 = 0 is σ 1 = 1 / 2 + ( 1 / 2 ) 5 ≈ 1.618 . The prediction of the method is for a negative root between − 1 / a 3 ≈ − 2.080 and 0; a positive root between σ 1 ≈ 1.618 and χ = 3 ; and another positive root greater than χ = 3 . Indeed, the roots are x 1 ≈ − 0.607 , x 2 ≈ 1.932 , and x 3 ≈ 7.674 .

3.1.2.4 The case of a > 0 , b < 0 , and c < 0

Given that b is again negative, the roots (19) are again always real and always with opposite signs. The relevant one is α 1 > 0 . The corresponding root of (11) is χ = χ 1 = ( c / b ) − 1 + 1 − 3 b / c 2 > − c / b > 0 . The curve a x 3 + 1 with a = 0 intersects in the first quadrant the curve − b x 2 − c x with b < 0 and c > 0 at point σ 1 = [ − c / ( 2 b ) ] 1 + 1 − 4 b / c 2 , which is the bigger root of the equation − b x 2 − c x − 1 = 0 .

Then the roots of the cubic equation are as follows. There is always one negative root between min − 1 / a 3 , − c / b and max − 1 / a 3 , − c / b . If a > α 1 , then the other two roots are complex. If a = α 1 , then, in addition to the negative root, there is a positive double root at χ . If a < α 1 , the roots are one negative root between min − 1 / a 3 , − c / b and max − 1 / a 3 , − c / b ; one positive root between σ 1 and χ ; and another positive root greater than χ (Figure 4c).

As an example, consider the equation

(41) 2 x 3 − x 2 − x + 1 = 0 .

The roots (19) are α 2 = − 5 / 27 (irrelevant) and α 1 = 1 . The corresponding loci of the double roots are χ 2 = − 3 (irrelevant) and χ = χ 1 = 1 . The bigger root of − b x 2 − c x − 1 = 0 is σ 1 = ( 1 / 2 ) − 1 + 5 ≈ 0.618 . Also − 1 / a 3 ≈ − 0.794 > − c / b = − 1 . The given a = 2 is greater than α 1 = 1 , thus the equation must have one negative root between c / b = − 1 and − 1 / a 3 ≈ − 0.794 and two complex roots. This is so indeed: the roots are x 1 ≈ − 0.829 and x 2 , 3 ≈ 0.665 ± 0.401 i .

The equation

(42) 1 2 x 3 − x 2 − x + 1 = 0

is another example chosen so that one again has α 2 = − 5 / 27 (irrelevant), α 1 = 1 , χ 2 = − 3 (irrelevant), χ = χ 1 = 1 , σ 1 = ( 1 / 2 ) − 1 + 5 ≈ 0.618 . This time − 1 / a 3 ≈ − 1.260 < − c / b = − 1 . The given a = 1 / 2 is now smaller than α 1 = 1 , thus the equation must have one negative root between − 1 / a 3 ≈ − 1.260 and c / b = − 1 and two positive roots – one between σ 1 ≈ 0.618 and χ = 1 and another one greater than χ = 1 . The roots are x 1 ≈ − 1.170 , x 2 ≈ 0.689 , and x 3 ≈ 2.481 – in their predicted bounds.

3.1.2.5 The four cases with a < 0

The analysis of these four cases is completely analogous as there is symmetry (reflection with respect to the ordinate) between them and the four cases already studied (one only needs to replace c by − c when a is replaced by − a ).

That is, the case of a < 0 , b > 0 , and c > 0 is analogous to the case of a > 0 , b > 0 , and c < 0 (Figure 4a); the case of a < 0 , b > 0 , and c < 0 is a complicated case analogous to the case of a > 0 , b > 0 , and c > 0 (Figure 3); the case of a < 0 , b < 0 , and c > 0 is analogous to the case of a > 0 , b < 0 , and c < 0 (Figure 4c); and, finally, the case of a < 0 , b < 0 , and c < 0 is analogous to the case of a > 0 , b < 0 , and c > 0 (Figure 4b).

3.2 An example with equation of degree 5

Consider the quintic equation

(43) − 1 12 x 5 + 1 4 x 4 + 5 12 x 3 − 5 4 x 2 − 1 3 x + 1 = 0

and split as follows

(44) − 1 12 x 5 + 1 = − 1 4 x 4 − 5 12 x 3 + 5 4 x 2 + 1 3 x .

Setting the discriminant of the quintic

(45) α x 5 + 1 4 x 4 + 5 12 x 3 − 5 4 x 2 − 1 3 x + 1

equal to zero results in an equation of degree 4,

(46) 3 125 α 4 + 9 78 851 972 α 3 + 1 05 19 165 1 86 624 α 2 − 11 49 707 11 19 744 α − 76 507 29 85 984 = 0 ,

the roots of which are α 1 ≈ − 0.015 , α 2 ≈ − 0.089 , α 3 ≈ − 0.24 , and α 4 ≈ 0.024 . As the given a is − 1 / 12 ≈ − 0.083 < 0 , the root α 4 > 0 is irrelevant. One also has α 3 < α 2 < a < α 1 . The double roots χ j , associated with α j , are the roots of the equation

(47) 1 4 x 4 + 2 × 5 12 x 3 + 3 × − 5 4 x 2 + 4 × − 1 3 x + 5 × 1 = 0 ,

namely, χ 1 ≈ 1.192 , χ 2 ≈ 2.398 , χ 3 ≈ − 1.229 , and the irrelevant χ 4 is − 5.695 approximately.

Next, solve the equations

(48) 1 4 x 4 + 5 12 x 3 − 5 4 x 2 − 1 3 x + 1 = 0

to determine the points at which the curve a x 5 + 1 with a = 0 intersects the curve − ( 1 / 4 ) x 4 − ( 5 / 12 ) x 3 + ( 5 / 4 ) x 2 + ( 1 / 3 ) x . The solutions are σ 1 ≈ − 0.965 , σ 2 ≈ − 3.028 , and σ 3 , 4 ≈ 1.163 ± 0.124 i .

The prediction of the model is for one negative root between σ 2 ≈ − 3.028 and χ 3 ≈ − 1.229 ; one negative root between χ 3 ≈ − 1.229 and σ 1 ≈ − 0.965 ; one positive root between 0 and χ 1 ≈ 1.192 ; one positive root between χ 1 ≈ 1.192 and χ 2 ≈ 2.398 ; and one positive root bigger than χ 2 ≈ 2.398 (Figure 5).

$Figure 5 The quintic equation − 1 12 x 5 + 1 4 x 4 + 5 12 x 3 − 5 4 x 2 − 1 3 x + 1 = 0 -\hspace{-0.25em}\tfrac{1}{12}{x}^{5}+\tfrac{1}{4}{x}^{4}+\tfrac{5}{12}{x}^{3}-\tfrac{5}{4}{x}^{2}-\tfrac{1}{3}x+1=0 .$

Figure 5

The quintic equation − 1 12 x 5 + 1 4 x 4 + 5 12 x 3 − 5 4 x 2 − 1 3 x + 1 = 0 .

The roots are x 1 = − 2 , x 2 = − 1 , x 3 = 1 , x 4 = 2 , and x 5 = 3 – within the predicted bounds.

3.3 Recursive application of the method. An example with equation of degree 7

Consider the equation of degree 7

(49) 16 3 x 7 − 52 3 x 6 + 14 3 x 5 + 77 3 x 4 − 77 6 x 3 − 28 3 x 2 + 17 6 x + 1 = 0

and see Figure 6.

$Figure 6 Recursive application of the method for the equation 16 3 x 7 − 52 3 x 6 + 14 3 x 5 + 77 3 x 4 − 77 6 x 3 − 28 3 x 2 + 17 6 x + 1 = 0 \tfrac{16}{3}{x}^{7}-\tfrac{52}{3}{x}^{6}+\tfrac{14}{3}{x}^{5}+\tfrac{77}{3}{x}^{4}-\tfrac{77}{6}{x}^{3}-\tfrac{28}{3}{x}^{2}+\tfrac{17}{6}x+1=0 . (a) The original equation of seventh degree is split as λ ( x ) − μ ( x ) = 0 \lambda (x)-\mu (x)=0 with λ ( x ) = 16 3 x 7 + 1 \lambda (x)=\tfrac{16}{3}{x}^{7}+1 and μ ( x ) = − 17 6 x ( − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 ) \mu (x)=\left-\tfrac{17}{6}x\left(-\tfrac{104}{17}{x}^{5}+\tfrac{28}{17}{x}^{4}+\tfrac{154}{17}{x}^{3}-\tfrac{77}{17}{x}^{2}-\tfrac{56}{17}x\left+1\right) . An equation of degree 5, μ ˆ ( x ) = − 104 17 x 5 + 28 17 x 4 \hat{\mu }(x)=-\tfrac{104}{17}{x}^{5}+\tfrac{28}{17}{x}^{4} + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 = 0 +\tfrac{154}{17}{x}^{3}-\tfrac{77}{17}{x}^{2}-\tfrac{56}{17}x+1=0 , results. (b) To find the roots of the resulting equation of degree 5, μ ˆ ( x ) = − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 = 0 \hat{\mu }(x)=-\tfrac{104}{17}{x}^{5}+\tfrac{28}{17}{x}^{4}+\tfrac{154}{17}{x}^{3}-\tfrac{77}{17}{x}^{2}-\tfrac{56}{17}x+1=0 , perform the further split μ ˆ ( x ) = ρ ( x ) − ν ( x ) \hat{\mu }(x)=\rho (x)-\nu (x) with ρ ( x ) = − 104 17 x 5 + 1 \rho (x)=-\tfrac{104}{17}{x}^{5}+1 and ν ( x ) = − 28 17 x 4 − 154 17 x 3 + 77 17 x 2 + 56 17 x \nu (x)=-\tfrac{28}{17}{x}^{4}-\tfrac{154}{17}{x}^{3}+\tfrac{77}{17}{x}^{2}+\tfrac{56}{17}x . (c) Since α 2 < − 104 17 < α 1 {\alpha }_{2}\lt -\tfrac{104}{17}\lt {\alpha }_{1} , the roots of μ ˆ ( x ) = ρ ( x ) − ν ( x ) = − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 = 0 \hat{\mu }(x)=\rho (x)-\nu (x)=-\tfrac{104}{17}{x}^{5}+\tfrac{28}{17}{x}^{4}+\tfrac{154}{17}{x}^{3}-\tfrac{77}{17}{x}^{2}-\tfrac{56}{17}x+1=0 are as follows: one negative root smaller than χ 2 {\chi }_{2} ; another negative root between χ 2 {\chi }_{2} and σ 2 {\sigma }_{2} ; a positive root smaller than − ( 104 / 17 ) − 1 / 7 -\hspace{-0.25em}{(104/17)}^{-1/7} ; and two complex roots.$

Figure 6

Recursive application of the method for the equation 16 3 x 7 − 52 3 x 6 + 14 3 x 5 + 77 3 x 4 − 77 6 x 3 − 28 3 x 2 + 17 6 x + 1 = 0 . (a) The original equation of seventh degree is split as λ ( x ) − μ ( x ) = 0 with λ ( x ) = 16 3 x 7 + 1 and μ ( x ) = − 17 6 x ( − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 ) . An equation of degree 5, μ ˆ ( x ) = − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 = 0 , results. (b) To find the roots of the resulting equation of degree 5, μ ˆ ( x ) = − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 = 0 , perform the further split μ ˆ ( x ) = ρ ( x ) − ν ( x ) with ρ ( x ) = − 104 17 x 5 + 1 and ν ( x ) = − 28 17 x 4 − 154 17 x 3 + 77 17 x 2 + 56 17 x . (c) Since α 2 < − 104 17 < α 1 , the roots of μ ˆ ( x ) = ρ ( x ) − ν ( x ) = − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 = 0 are as follows: one negative root smaller than χ 2 ; another negative root between χ 2 and σ 2 ; a positive root smaller than − ( 104 / 17 ) − 1 / 7 ; and two complex roots.

In order to find the number of positive and negative roots of this equation, together with their bounds, one cannot consider setting its discriminant equal to zero as the resulting equation for the unknown α (which replaces the coefficient 16/3 of x 7 ) will be of degree 6 and not solvable analytically. One has to proceed by applying the method twice. First, rewrite the given equation p 7 ( x ) = 0 as λ ( x ) − μ ( x ) = 0 with

(50) λ ( x ) = 16 3 x 7 + 1 ,

(51) μ ( x ) = − 17 6 x − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 ≡ − 17 6 x μ ˆ ( x ) .

The only real root of λ ( x ) = 0 is λ 1 = − 3 / 16 7 ≈ − 0.787 . The roots of μ ( x ) = 0 are those of μ ˆ ( x ) = 0 and zero. If the number of positive and negative ones among those can be determined, together with their bounds, then they can be used to attempt to determine the number of positive and negative roots (and their respective bounds) of the original equation.

Consider next the resulting equation of degree 5, μ ˆ ( x ) = 0 , that is

(52) − 104 17 x 5 + 28 17 x 4 + 154 17 x 3 − 77 17 x 2 − 56 17 x + 1 = 0 ,

and split further: μ ˆ ( x ) = ρ ( x ) − ν ( x ) = 0 , where:

(53) ρ ( x ) = − 104 17 x 5 + 1 ,

(54) ν ( x ) = − 28 17 x 4 − 154 17 x 3 + 77 17 x 2 + 56 17 x .

The only real root of ρ ( x ) = 0 is ρ 1 = − 17 / 104 5 ≈ 0.696 . The equation ν ( x ) = 0 is of degree 4 and its roots are ν 1 ≈ − 5.908 , which is not shown in Figure 6b to avoid scaling of the graph, ν 2 ≈ − 0.412 , ν 3 = 0 , and ν 4 ≈ 0.821 .

Considering the quintic discriminant of α x 5 + ( 28 / 17 ) x 4 + ( 154 / 17 ) x 3 − ( 77 / 17 ) x 2 − ( 56 / 17 ) x + 1 and setting it to zero result in an equation of degree 4 for the unknown α the roots of which are α 1 ≈ − 3.955 , α 2 ≈ − 10.021 and the irrelevant ones α 3 ≈ 0.072 and α 4 ≈ 60.743 (given that the coefficient of x 5 in ρ ( x ) is − 104 / 17 ≈ − 6.118 < 0 and α 3 , 4 are positive).

The corresponding loci of the double roots of the equations α j x 5 + ( 28 / 17 ) x 4 + ( 154 / 17 ) x 3 − ( 77 / 17 ) x 2 − ( 56 / 17 ) x + 1 = 0 , j = 1 , 2 , 3 , 4 , coincide with the roots of the equation

(55) 2 17 x 4 + 2 × 154 17 x 3 + 3 × − 77 17 x 2 + 4 × − 56 17 x + 5 × 1 = 0

and are χ 1 ≈ 1.081 , χ 2 ≈ − 0.753 , and the irrelevant χ 3 ≈ − 11.648 and χ 4 ≈ 0.320 .

The intersection points of the curve α x 5 + 1 for which α = 0 with the curve ν ( x ) are the roots of the equation ( 28 / 17 ) x 4 + ( 154 / 17 ) x 3 − ( 77 / 17 ) x 2 − ( 56 / 17 ) x + 1 = 0 , namely, σ 1 ≈ − 5.905 , σ 2 ≈ − 0.559 , σ 3 ≈ 0.261 , and σ 4 ≈ 0.704 .

The coefficient of x 5 in ρ ( x ) is − 104 / 17 ≈ − 6.118 < 0 and this is greater than α 2 ≈ − 10.021 and smaller than α 1 ≈ − 3.955 . Therefore, the roots of the equation μ ( x ) = − [ ( 17 x ) / 6 ] [ − ( 104 / 17 ) x 5 + ( 28 / 17 ) x 4 + ( 154 / 17 ) x 3 − ( 77 / 17 ) x 2 − ( 56 / 17 ) x + 1 ] = 0 are as follows: negative root μ 1 smaller than χ 2 ≈ − 0.753 ; negative root μ 2 between χ 2 ≈ − 0.753 and σ 2 ≈ − 0.559 ; μ 3 = 0 ; positive root μ 4 smaller than ρ 1 = − ( 104 / 17 ) − 1 / 7 ≈ 0.547 ; and two complex roots μ 5 , 6 . The actual roots are μ 1 ≈ − 1.140 , μ 2 ≈ − 0.612 , μ 3 = 0 , μ 4 ≈ 0.259 , and μ 5 , 6 ≈ 0.881 ± 0.359 i – within their predicted bounds.

Returning with the roots of μ ( x ) = 0 to the first split, p 7 ( x ) = λ ( x ) − μ ( x ) = 0 , one can determine the following (Figure 6a). The given equation p 7 ( x ) = 0 has one negative root x 1 between μ 1 and λ 1 = − 3 / 16 7 ≈ − 0.787 . But given that μ 1 < χ 2 ≈ − 0.753 , then all that can be said about the root x 1 is that it must be smaller than λ 1 ≈ − 0.787 . The actual root of the equation is x 1 = − 1 . There can be no roots between λ 1 ≈ − 0.787 and μ 2 as the function μ ( x ) is negative, while the function λ ( x ) is positive there. Given that μ 2 is between χ 2 ≈ − 0.753 and σ 2 ≈ − 0.559 , then there could be no roots between λ 1 ≈ − 0.787 and χ 2 ≈ − 0.753 . Next, consider the sub-interval between μ 2 and 0. As μ 2 is between χ 2 ≈ − 0.753 and σ 2 ≈ − 0.559 and given that μ ( x ) is of degree 5 and thus it can have up to four extremal points, then there could be either zero or two negative roots of the original equation between and σ 2 ≈ − 0.559 and the origin. The actual roots of the equation there are two: x 2 = − 1 / 2 and x 3 = − 1 / 4 . Given that between μ 3 = 0 and μ 4 , the function μ ( x ) is negative, while the function λ ( x ) is positive, then there could be no intersection of these two curves in this interval and the equation p 7 ( x ) = 0 cannot have roots in it. However, given that μ 4 is between 0 and ρ 1 = − ( 104 / 17 ) − 1 / 7 ≈ 0.547 , what can be said about the positive roots of this equation is that there is either two or four of them. The actual roots are x 4 = 1 / 2 , x 5 = 1 , x 6 = 3 / 2 , and x 7 = 2 .

A step-by-step algorithm for the recursive application of the split (2) is given in Section 8.

4 Using the split (2). An example with equation of degree 9

Consider the following example as an equation of degree above 5 and up to and including 9:

(56) x 9 + 1 2 x 8 − 7 x 7 − 2 x 6 + 9 x 5 − x 4 − 2 x 3 + 13 x 2 + 14 x − 24 = 0 .

Using the split (2), this equation can be written as f ( x ) − g ( x ) = 0 with

(57) f ( x ) = x 4 + 1 2 x 3 − 7 x 2 − 2 x + 9 x 5 ,

(58) g ( x ) = x 4 + 2 x 3 − 13 x 2 − 14 x + 24 .

The roots of the two equations f ( x ) = 0 and g ( x ) = 0 can be determined analytically.

For f ( x ) = 0 , these are f 1 ≈ − 2.416 , f 2 ≈ − 1.458 , f 3 , 4 , 5 , 6 , 7 = 0 (zero is a quintuple root), f 8 ≈ 1.145 , and f 9 ≈ 2.230 . The origin is a saddle for f ( x ) . The first four derivatives of f ( x ) at 0 are zero, while the fifth one is positive. Thus, f ( x ) enters through the origin into the first quadrant from the third. The function f ( x ) has two negative roots and two positive ones. Furthermore, when x → − ∞ , f ( x ) → − ∞ , while when x → ∞ , f ( x ) → ∞ . It is essential to note that f ( x ) can have up to four non-zero extremal points. Namely, f ′ ( x ) = 9 x 8 + 4 x 7 − 49 x 6 − 12 x 5 + 45 x 4 and setting this to zero gives the following extremal points: a quadruple 0, together with the points: − 2.179 , − 1.210 , 0.952 , and 1.992.

The roots of g ( x ) = 0 are g 1 = − 4 , g 2 = − 2 , g 3 = 1 , and g 4 = 3 . At zero, one has g ( 0 ) = 24 > 0 . The function g ( x ) tends to + ∞ when x → ± ∞ . It also has two positive roots and two negative ones. The function g ( x ) can have up to three extremal points. These are the roots of the equation g ′ ( x ) = 4 x 3 + 6 x 2 − 26 x − 14 = 0 , namely, the points − 3.193 , − 0.5 , and 2.193.

This very simple analysis allows the graphs of f ( x ) and g ( x ) to be easily sketched (Figure 7) and from the graph one can infer the following for the roots x i of the given equation. There can be no roots smaller than g 1 = − 4 as in that region f ( x ) and g ( x ) have different signs and thus cannot intersect. There is one negative root between g 1 = − 4 and f 1 ≈ − 2.416 . Given the number of extremal points of f ( x ) , it is not possible to have more negative roots in this sub-interval. There can be no roots between f 1 ≈ − 2.416 and g 2 = 2 . There is one negative root between g 2 = − 2 and f 2 ≈ − 1.458 . Again, due to the number of extremal points of f ( x ) , there can be no other negative roots in this sub-interval. There are no roots between f 2 ≈ − 1.458 and zero. There is one and no more positive roots between 0 and g 3 = 1 . There can be no roots between g 3 = 1 and f 8 ≈ 1.145 . Between f 8 ≈ 1.145 and f 9 ≈ 2.230 , there can be either zero or two positive roots. There can be no roots between f 9 ≈ 2.230 and g 4 = 3 . As f ( x ) grows faster than g ( x ) , there can be no roots greater than g 4 = 3 either. The biggest positive root is therefore smaller than f 9 ≈ 2.230 . All roots are locked between g 1 = − 4 and f 9 ≈ 2.230 .

$Figure 7 The equation x 9 + 1 2 x 8 − 7 x 7 − 2 x 6 + 9 x 5 − x 4 − 2 x 3 + 13 x 2 + 14 x − 24 = 0 {x}^{9}+\tfrac{1}{2}{x}^{8}-7{x}^{7}-2{x}^{6}+9{x}^{5}-{x}^{4}-2{x}^{3}+13{x}^{2}+14x-24=0 .$

Figure 7

The equation x 9 + 1 2 x 8 − 7 x 7 − 2 x 6 + 9 x 5 − x 4 − 2 x 3 + 13 x 2 + 14 x − 24 = 0 .

The actual roots of the equation are x 1 ≈ − 2.426 , x 2 ≈ − 1.591 , x 3 ≈ 0.948 , x 4 ≈ 1.388 , x 5 ≈ 2.203 , x 6 , 7 ≈ − 0.917 ± 0.837 i , and x 8 , 9 ≈ 0.406 ± 1.107 i .

For comparison, the Descartes rule of signs provides for 1, 3, or 5 positive roots and for 0, 2, or 4 negative roots.

The Lagrange bound provides that all roots are between ± max 1 , ∑ i = 0 n − 1 | a i / a n | = ± 72.5 . Cauchy’s theorem provides a stricter bound on all roots: they are locked between ± 1 + max 0 ≤ k ≤ n − 1 | a k | = ± 25 .

A step-by-step algorithm for the recursive application of the split (5) is given in Section 7.

5 Application of the method to a realistic scientific example involving a quartic equation with varying coefficients

Constantin and Ivanov [8] studied the nonlinear equations of motion for equatorial wave-current interactions in the physically realistic model of azimuthal two-dimensional non-viscous flows with piecewise constant vorticity in a two-layer fluid with a flat bed and a free surface. The derived Hamiltonian formulation for the nonlinear governing equations yields, upon linearization, a homogeneous system (no sources are being considered) of four simultaneous equations – the linearized Hamilton equations of motion. Non-trivial solution of this system is sought and the corresponding requirement for vanishing determinant of the homogenous linear system leads to a quartic equation for the wave speed x relative to the maximum speed of the Equatorial Undercurrent. The two layers of the fluid, having the same wave number, have the same wave speed x and this x could be each one of the four possible roots of this quartic equation. In the regime of long waves (characterized by wavelengths in excess of 16 km), the resulting quartic equation is [8]:

(59) x 4 − ( γ 1 h 1 − γ h ) x 3 − [ g ( h + h 1 ) + γ γ 1 h h 1 ] x 2 + g h h 1 ( γ 1 − γ ) x + r g 2 h h 1 1 + r = 0 ,

where all quantities are dimensionless and as follows: γ is the constant vorticity of the lower layer, γ 1 is the constant vorticity of the upper layer, h is the depth of the lower layer, h 1 is the depth of the upper layer, g is the acceleration due to gravity, and r = ( ρ − ρ 1 ) / ρ 1 is the relative density difference between the two layers (with ρ and ρ 1 being the densities of the lower and the upper layer, respectively).

As pointed out by the authors [8],“the intricacy of the coefficients prevents (one) from using the explicit quartic formulae to gain insight into the location of the roots” and they proceed by reducing the solution of the quartic by addressing the resolvent cubic. Even for this, it is “a daunting task to use Cardano’s formulae to identify a real root of the cubic” [8] and the authors exploit the available structure to gain insight into the location of the roots and avoid the computational approach. The existence of four real roots of the above quartic equation is implied.

The typical values of the physical parameters involved are as follows [8]: h ≈ 8 , h 1 ≈ 0.24 , γ h ≈ 2 , γ 1 h 1 ≈ − 3 , g ≈ 2 × 10 4 , and r ≈ 10 − 3 . Thus, within this range of parameters, the resulting quartic equation

(60) x 4 + a x 3 + b x 2 + c x + d = 0

has a = − ( γ 1 h 1 − γ h ) > 0 , b = − [ g ( h + h 1 ) + γ γ 1 h h 1 ] < 0 , c = g h h 1 ( γ 1 − γ ) < 0 , and d = ( r g 2 h h 1 ) ( 1 + r ) − 1 > 0 .

As this is an equation with four “degrees of freedom” (the coefficients can be varied), it is suggestive to split them equally and, together with this, the quartic equation itself:

(61) x 2 ( x 2 + a x + b ) = − c x − d .

On the left-hand side, one has a quartic with zero as a double root and two more roots ρ 1 , 2 given by

(62) ρ 1 , 2 = 1 2 − a ± a 2 − 4 b .

Within the above range of model parameters, the roots ρ 1 , 2 are real (as b < 0 ).

On the right-hand side of (61) is a straight line and the possible wave speeds x are given by the intersection points of this straight line with the quartic x 2 ( x 2 + a x + b ) (Figure 8).

$Figure 8 (a) c 2 < γ 2 < c 0 < c < 0 {c}_{2}\lt {\gamma }_{2}\lt {c}_{0}\lt c{\boldsymbol{\lt }}{\bf{0}} . (i) When − d < − δ 3 -\hspace{-0.25em}d\lt -{\delta }_{3} , there are no real roots. (ii) When − δ 3 ≤ − d < − δ 1 -\hspace{-0.25em}{\delta }_{3}\le -d\lt -{\delta }_{1} , there are two negative roots x 1 , 2 {x}_{1,2} such that ξ 2 ( 1 ) < x 2 ≤ χ 3 {\xi }_{2}^{(1)}\lt {x}_{2}\le {\chi }_{3} and χ 3 ≤ x 1 < ξ 1 ( 1 ) {\chi }_{3}\le {x}_{1}\lt {\xi }_{1}^{(1)} . (iii) When − δ 1 ≤ − d < 0 -\hspace{-0.25em}{\delta }_{1}\le -d\lt 0 , there are four real roots – two negative: λ 3 < x 4 ≤ ξ 2 ( 1 ) {\lambda }_{3}\lt {x}_{4}\le {\xi }_{2}^{(1)} and ξ 1 ( 1 ) ≤ x 3 < λ 2 {\xi }_{1}^{(1)}\le {x}_{3}\lt {\lambda }_{2} and two positive: 0 < x 2 ≤ min { − d / c , χ 1 } 0\lt {x}_{2}\le \hspace{.25em}\min \hspace{.25em}\{-d/c,{\chi }_{1}\} and max { − d / c , χ 1 } ≤ x 1 < λ 1 \{-d/c,{\chi }_{1}\}\le {x}_{1}\lt {\lambda }_{1} . Analysis based on solving quadratic equations only. (i) When − d < c ρ 1 -\hspace{-0.25em}d\lt c{\rho }_{1} , there are no real roots, there are two negative roots x 1 , 2 > ρ 2 {x}_{1,2}\gt {\rho }_{2} , or there are four real roots: ρ 2 < x 3 , 4 < 0 {\rho }_{2}\lt {x}_{3,4}\lt 0 and 0 < x 1 , 2 < ρ 1 0\lt {x}_{1,2}\lt {\rho }_{1} . (ii) When c ρ 1 ≤ − d < 0 c{\rho }_{1}\le -d\lt 0 , there are four real roots – two negative: ρ 2 < x 3 , 4 < 0 {\rho }_{2}\lt {x}_{3,4}\lt 0 and two positive: 0 < x 2 ≤ − d / c 0\lt {x}_{2}\le -d/c , and x 1 ≥ ρ 1 {x}_{1}\ge {\rho }_{1} . (b) c 2 < γ 2 < c < c 0 < 0 {{\boldsymbol{c}}}_{2}\lt {\gamma }_{2}\lt c\lt {c}_{0}\lt 0 . (i) When − d < − δ 1 -\hspace{-0.25em}d\lt -{\delta }_{1} , there are no real roots. (ii) When − δ 1 ≤ − d < − δ 3 -\hspace{-0.25em}{\delta }_{1}\le -d\lt -{\delta }_{3} , there are two positive roots: ξ 2 ( 3 ) < x 2 ≤ χ 1 {\xi }_{2}^{(3)}\lt {x}_{2}\le {\chi }_{1} and χ 1 ≤ x 1 < ξ 1 ( 3 ) {\chi }_{1}\le {x}_{1}\lt {\xi }_{1}^{(3)} . (iii) When − δ 3 ≤ − d < 0 -\hspace{-0.25em}{\delta }_{3}\le -d\lt 0 , there are four real roots – two negative: λ 3 < x 4 ≤ χ 3 {\lambda }_{3}\lt {x}_{4}\le {\chi }_{3} and χ 3 ≤ x 3 < λ 2 {\chi }_{3}\le {x}_{3}\lt {\lambda }_{2} and two positive: 0 < x 2 ≤ 0\lt {x}_{2}\hspace{0.25em}\le min { − d / c , χ 1 } \{-d/c,{\chi }_{1}\} and max { − d / c , χ 1 } ≤ x 1 < λ 1 \{-d/c,{\chi }_{1}\}\le {x}_{1}\lt {\lambda }_{1} . Analysis based on solving quadratic equations only. (i) When − d < c ρ 1 -\hspace{-0.25em}d\lt c{\rho }_{1} , there are no real roots, there are two positive roots x 1 , 2 < ρ 1 {x}_{1,2}\lt {\rho }_{1} , or there are four real roots: ρ 2 < x 3 , 4 < 0 {\rho }_{2}\lt {x}_{3,4}\lt 0 and 0 < x 1 , 2 < ρ 1 0\lt {x}_{1,2}\lt {\rho }_{1} . (ii) When c ρ 1 ≤ − d < 0 c{\rho }_{1}\le -d\lt 0 , there are four real roots – two negative: ρ 2 < x 3 , 4 < 0 {\rho }_{2}\lt {x}_{3,4}\lt 0 and two positive: 0 < x 2 ≤ − d / c 0\lt {x}_{2}\le -d/c , and x 1 ≥ ρ 1 {x}_{1}\ge {\rho }_{1} . (c) c 2 < c < γ 2 < c 0 < 0 {{\boldsymbol{c}}}_{2}\lt c\lt {\gamma }_{2}\lt {c}_{0}\lt 0 . (i) When − d < − δ 1 -\hspace{-0.25em}d\lt -{\delta }_{1} , there are no real roots. (ii) When − δ 1 ≤ − d < c ρ 1 -\hspace{-0.25em}{\delta }_{1}\le -d\lt c{\rho }_{1} , there are two positive roots: x 2 ≤ χ 1 {x}_{2}\le {\chi }_{1} and χ 1 ≤ x 1 < ρ 1 {\chi }_{1}\le {x}_{1}\lt {\rho }_{1} . (iii) When c ρ 1 ≤ − d < 0 c{\rho }_{1}\le -d\lt 0 , there are again two positive roots: x 2 ≤ min { − d / c , χ 1 } {x}_{2}\le \hspace{.25em}\min \{-d/c,{\chi }_{1}\} and ρ 1 ≤ x 1 < λ 1 {\rho }_{1}\le {x}_{1}\lt {\lambda }_{1} . Analysis based on solving quadratic equations only. (i) When − d < c ρ 1 -\hspace{-0.25em}d\lt c{\rho }_{1} , there are either no real roots or there are two positive roots x 1 , 2 < ρ 1 {x}_{1,2}\lt {\rho }_{1} . (ii) When c ρ 1 ≤ − d < 0 c{\rho }_{1}\le -d\lt 0 , there are two positive roots: x 2 ≤ − d / c {x}_{2}\le -d/c and x 1 ≥ ρ 1 {x}_{1}\ge {\rho }_{1} . (d) c < c 2 < γ 2 < c 0 < 0 {\boldsymbol{c}}\lt {c}_{2}\lt {\gamma }_{2}\lt {c}_{0}\lt 0 . If − d -\hspace{-0.25em}d is allowed to take positive values, then this would be a different situation. However, as − d < 0 -\hspace{-0.25em}d\lt 0 , one has the same possibilities as those for − d < 0 -\hspace{-0.25em}d\lt 0 on (c).$

Figure 8

(a) c 2 < γ 2 < c 0 < c < 0 . (i) When − d < − δ 3 , there are no real roots. (ii) When − δ 3 ≤ − d < − δ 1 , there are two negative roots x 1 , 2 such that ξ 2 ( 1 ) < x 2 ≤ χ 3 and χ 3 ≤ x 1 < ξ 1 ( 1 ) . (iii) When − δ 1 ≤ − d < 0 , there are four real roots – two negative: λ 3 < x 4 ≤ ξ 2 ( 1 ) and ξ 1 ( 1 ) ≤ x 3 < λ 2 and two positive: 0 < x 2 ≤ min { − d / c , χ 1 } and max { − d / c , χ 1 } ≤ x 1 < λ 1 . Analysis based on solving quadratic equations only. (i) When − d < c ρ 1 , there are no real roots, there are two negative roots x 1 , 2 > ρ 2 , or there are four real roots: ρ 2 < x 3 , 4 < 0 and 0 < x 1 , 2 < ρ 1 . (ii) When c ρ 1 ≤ − d < 0 , there are four real roots – two negative: ρ 2 < x 3 , 4 < 0 and two positive: 0 < x 2 ≤ − d / c , and x 1 ≥ ρ 1 . (b) c 2 < γ 2 < c < c 0 < 0 . (i) When − d < − δ 1 , there are no real roots. (ii) When − δ 1 ≤ − d < − δ 3 , there are two positive roots: ξ 2 ( 3 ) < x 2 ≤ χ 1 and χ 1 ≤ x 1 < ξ 1 ( 3 ) . (iii) When − δ 3 ≤ − d < 0 , there are four real roots – two negative: λ 3 < x 4 ≤ χ 3 and χ 3 ≤ x 3 < λ 2 and two positive: 0 < x 2 ≤ min { − d / c , χ 1 } and max { − d / c , χ 1 } ≤ x 1 < λ 1 . Analysis based on solving quadratic equations only. (i) When − d < c ρ 1 , there are no real roots, there are two positive roots x 1 , 2 < ρ 1 , or there are four real roots: ρ 2 < x 3 , 4 < 0 and 0 < x 1 , 2 < ρ 1 . (ii) When c ρ 1 ≤ − d < 0 , there are four real roots – two negative: ρ 2 < x 3 , 4 < 0 and two positive: 0 < x 2 ≤ − d / c , and x 1 ≥ ρ 1 . (c) c 2 < c < γ 2 < c 0 < 0 . (i) When − d < − δ 1 , there are no real roots. (ii) When − δ 1 ≤ − d < c ρ 1 , there are two positive roots: x 2 ≤ χ 1 and χ 1 ≤ x 1 < ρ 1 . (iii) When c ρ 1 ≤ − d < 0 , there are again two positive roots: x 2 ≤ min { − d / c , χ 1 } and ρ 1 ≤ x 1 < λ 1 . Analysis based on solving quadratic equations only. (i) When − d < c ρ 1 , there are either no real roots or there are two positive roots x 1 , 2 < ρ 1 . (ii) When c ρ 1 ≤ − d < 0 , there are two positive roots: x 2 ≤ − d / c and x 1 ≥ ρ 1 . (d) c < c 2 < γ 2 < c 0 < 0 . If − d is allowed to take positive values, then this would be a different situation. However, as − d < 0 , one has the same possibilities as those for − d < 0 on (c).

For this particular split, one can denote the free term d by δ and vary δ in such a way that the original quartic is endowed with a double root χ . As earlier, two equations are satisfied in this case: χ 2 ( χ 2 + a χ + b ) = − c χ − d and [ χ 2 ( χ 2 + a χ + b ) ] ′ = ( − c χ − d ) ′ . The latter is simply

(63) 4 χ 3 + 3 a χ 2 + 2 b χ + c = 0 .

This cubic equation is referred to as the first auxiliary equation (to make a distinction from the resolvent cubic equation). The discriminant Δ of the first auxiliary equation is

(64) Δ = − 432 c 2 − 108 a ( a 2 − 4 b ) c + 36 a 2 b 2 − 128 b 3

and it is non-negative for c, such that c 2 ≤ c ≤ c 1 , where

(65) c 1 , 2 = c 0 ± 3 72 ( 3 a 2 − 8 b ) 3 ,

with c 0 = ( 1 / 2 ) a ( b − a 2 / 4 ) .

In this case, Eq. (63) will have three real roots: χ 1 , χ 2 , and χ 3 . Then, using (61), one gets:

(66) δ i = − χ i 4 − a χ i 3 − b χ i 2 − c χ i for each i = 1 , 2 , 3 .

Therefore, the non-negativity of Δ leads to three possible tangents − c x − δ i to x 2 ( x 2 + a x + b ) – at each of the three possible double roots χ i . The original quartic Eq. (59) can thus have no real roots, it can have two real roots, or it can have four real roots.

If, on the other hand, Δ is negative, then the first auxiliary Eq. (63) has only one real root χ 1 , and there is only one straight line − c x − δ 1 that can be tangent to x 2 ( x 2 + a x + b ) . In such case, the quartic Eq. (59) can have either no real roots or it can have two real roots.

One can have either − δ 3 < − δ 1 or − δ 1 < − δ 3 (note that, in view of c < 0 , one always has − δ 2 > 0 and, also, − δ 1 < 0 ) (Figure 8). Which of these two situations is realized depends on the slope − c of the straight line − c x − d . The straight line − c 0 x − d 0 , with the above c 0 and with d 0 = ( 1 / 4 ) ( b − a 2 / 4 ) 2 , is tangent to x 2 ( x 2 + a x + b ) at two points, i.e. the equation x 4 + a x 3 + b x 2 + c 0 x + d 0 = 0 has two double roots. If the given c is greater than c 0 , then − δ 3 < − δ 1 (Figure 8a). If c < c 0 , then − δ 1 < − δ 3 (Figure 8b).

One should also note that − δ 3 can be positive or negative (Figure 8). There is an easy way to determine which of these two occurs: it suffices to see if the straight line − c x crosses x 2 ( x 2 + a x + b ) three times (in which case − δ 3 < 0 , Figure 8a and b) or once (in which case − δ 3 > 0 , Figure 8c). Namely, one has to study the second auxiliary equation

(67) x 3 + a x 2 + b x + c = 0 ,

the roots of which are denoted in Figure 8 by λ 1 , λ 2 , and λ 3 .

The discriminant of this equation is − 27 c 2 + ( − 4 a 3 + 18 a b ) c + b 2 ( a 2 − 4 b ) and it is non-negative for γ 2 ≤ c ≤ γ 1 , where

(68) γ 1 , 2 = − 2 27 a 3 + 1 3 a b + 2 27 ( a 2 − 3 b ) 3 .

When γ 2 ≤ c ≤ γ 1 , all λ ’s are real and thus − δ 3 < 0 . Otherwise, only λ 1 is real and thus − δ 3 > 0 .

In the case of three real roots χ i of the first auxiliary cubic equation (that is, the stationary points of the quartic being three), of the three tangents − c x − δ i to x 2 ( x 2 + a x + b ) at χ i , one does not intersect x 2 ( x 2 + a x + b ) further, while each of the remaining two tangents intersects x 2 ( x 2 + a x + b ) at two further points ξ 1 , 2 ( i ) . These can be easily found. If χ i is a double root of the quartic, then, using Viète’s formulae for the quartic, allows one to find that ξ 1 , 2 ( i ) are roots of the quadratic equation

(69) x 2 + ( a + 2 χ i ) x + 3 χ i 2 + 2 a χ i + b = 0 .

In the case of just one real root χ 1 of the first auxiliary cubic equation (that is, the quartic having a single stationary point), the tangent − c x − δ 1 to x 2 ( x 2 + a x + b ) at χ 1 does not intersect x 2 ( x 2 + a x + b ) further.

For the case of a > 0 , b < 0 , and c < 0 , one has c 2 ( a , b ) < γ 2 ( a , b ) < c 0 ( a , b ) < 0 < γ 1 ( a , b ) < c 1 ( a , b ) . Thus, there are four possible situations that can occur and all of these are shown in Figure 8. In Figure , one has c 2 < γ 2 < c 0 < c < 0 , Figure 8b is with c 2 < γ 2 < c < c 0 < 0 , Figure 8c is with c 2 < c < γ 2 < c 0 < 0 , and Figure 8d is with c < c 2 < γ 2 < c 0 < 0 .

The two auxiliary equations are cubic and, despite being significantly simpler than the resolvent cubic equation for any quartic, the application of Cardano’s formulae is still a daunting task. Assuming that the roots of the two auxiliary cubic equations are found, the full classification of the roots of the given quartic, together with the root isolation intervals, is presented in the captions of Figure 8. In addition, parallel analysis, based on solving quadratic equations only, is also presented in the captions of Figure 8, and this analysis yields remarkably good results.

6 A different perspective on the split (5)

Consider the general equation of degree n:

(70) x n + a n − 1 x n − 1 + a n − 2 x n − 2 + ⋯ + a 1 x + a 0 = 0 .

(Without loss of generality, the coefficient of the n th power of x can be considered equal to 1.)

The split

(71) x n + a 0 = − a n − 1 x n − 1 − a n − 2 x n − 2 − ⋯ − a 1 x

would allow the “propagation” of the curve of fixed shape x n vertically until one finds the tangent points of x n + a 0 with the curve − a n − 1 x n − 1 − a n − 2 x n − 2 − ⋯ − a 1 x . In this case, one will again have to solve equations of degree 1 smaller than that of the given equation as n − 1 is the highest power of a 0 in the discriminant Δ n of x n + a n − 1 x n − 1 + a n − 2 x n − 2 + ⋯ + a 1 x + a 0 . Then, in order to determine the number of positive and negative roots of the given equation and to also find their bounds, the given y-intercept a 0 will have to be compared to the real roots γ i of the equation Δ n ( γ ) = 0 in which a 0 has been replaced by γ and treated as an unknown, while all other a j ( j = 1 , 2 , … , n − 1 ) are as given. This is an equation of degree n − 1 and it is solvable analytically for n ≤ 5 . For n ≥ 6 , this split should be applied recursively. The idea is very similar to the one studied in detail in this paper.

7 Step-by-step algorithm for the application of the split (5)

(I) Equations of degree 5 or less

A single application of the method is needed for equations of degree up to and including five.

Note the sign of the coefficient, say a, of the highest power of x (say, n ≤ 5 ).
Replace the given coefficient a with α and treat α as a variable (the original equation is being varied).
Set the discriminant of the resulting polynomial equal to zero. This will be an equation of degree 1 less than that of the original equation.
Solve this equation and discard the complex roots. Also discard those roots that have sign opposite to that of the given coefficient a. Discard any repeated roots (those would yield the same χ ’s in step 6). Re-label the remaining roots in the ascending order: α 1 < α 2 < …
Remove the term a x n from the given equation and solve the remaining equation (this will also be an equation of degree 1 less than that of the original equation). Solve this equation and discard the complex roots. Re-label the remaining roots in the ascending order: σ 1 ≤ σ 2 ≤ …
In the given equation, multiply the term a x n by zero, the next term by one, the third term by two, and so on. The resulting equation is also of degree 1 less than that of the original equation. Solve this equation. The roots of this equation are the double roots χ i of the varied original equation and these correspond to the different α i .
From the roots χ i determined in the previous step, one has to discard those that correspond to the discarded real α in step 4. The χ ’s are “sifted” by the direct substitution of each of the χ ’s into the equation in step 2 with each of the retained α ’s: if an identity is obtained, the double root χ is retained. Re-label the retained χ ’s so that their indexes match the indexes of the corresponding α ’s.
In the original equation, remove the term a x n and the free term. The remaining equation is of order 1 less than that of the original equation and (at least) one of the roots of this equation is zero. It also has up to three real non-zero roots λ 1 ≤ λ 2 ≤ λ 3 .
This step is not necessary, but is very helpful: sketch the graphs of both parts of the split (5) and mark all χ ’s, σ ’s, and λ ’s. Mark the roots(s) of a x n + 1 = 0 . Also, draw the graphs of α i x + 1 .
Each of the graphs of α i x + 1 is tangent to r n − 1 ( x ) from the split (5) at the corresponding χ i . Find the “place” of the given coefficient a in the chain of inequalities, for example: α 1 < α 2 < a < α 3 .
If a is positive and bigger than all α ’s, then, if the original equation is of an odd degree, it has one real root which is negative and bigger than − 1 / a n . Otherwise, it does not have real roots. Similarly, if a is negative and smaller than all the α ’s: the only real root will be positive and smaller than − 1/ a n for an equation of an odd degree or there will be no real roots if the equation is of an even degree.
For each α “passed” by a (i.e. step 11 not being applicable), two roots appear on either side of the corresponding χ . Each of these roots is bounded on one side by this particular χ and, if they exist, by the σ ’s on either side of that χ . If there is no σ , then the other bound is the next double root χ . The λ ’s also serve as bounds on individual roots as, for every λ , the signs of the two parts of the split (5) are different on one side of this particular λ and the two parts of the split cannot cross for as long as another λ is reached in that direction. The root(s) of a x n + 1 could also serve as bounds on individual roots in the exact same way as the λ ’s – the two parts of the split (5) cannot cross if they have different signs.

(II) Equations of degree 6 or more

For equations of higher degrees, recursive application is needed.

Note the sign of the coefficient, say a, of the highest power of x (say n ≥ 6 ).
Remove from the equation the term with the highest power of x and the free term. For each of the remaining n − 1 terms, lower the power of x by one. Multiply the resulting equation by the reciprocal of the new free term.
For this new equation, apply steps 1 and 2 until a quintic equation is reached. There will be a different split of the type (5) each time the above is done.
Apply the method (I) for this quintic equation to find the isolation interval of each of the roots.
Return to the equation of degree 6 preceding the quintic equation. Use the roots of a x 6 + 1 and the isolation intervals obtained in step 5 to find the bounds for each of the roots of the equation of degree 6 by determining where a x 6 + 1 and the quintic have the same sign, as only in such subinterval a root of the equation of degree 6 can exist.
Note that if a x 6 + 1 and the quintic have the same sign in a given subinterval, the number of crossing points in that subinterval can be determined modulo an even number. There is a limit, however, on that even number – one has to factor in the fact that equation of degree 6 cannot have more than five stationary points, and thus there is a limit on the sign changes of the derivative and hence on the number of crossing points is such subinterval.
Use the obtained bounds in steps 5 and 6 to determine bounds for the roots of the equation of degree 7, preceding the equation of degree 6, by following the same procedure.
Repeat until the original equation is reached.

8 Step-by-step algorithm for the application of the split (2)

(I) Equations of degree 9 or less

A single application of the method is needed for equations of degree up to and including nine.

Select k consecutive terms from the polynomial ( k ≤ 5 ), starting with the term with the highest power of x, say n ( n ≤ 9 ). The “point” k of the split (2) should be chosen in such a way that one does not end up with equation of degree greater than four on the right-hand side of the split (2), otherwise the method will have to be applied twice. For example, if n = 9 , the obvious choice would be k = 5 ; for n = 8 , one should chose k = 4 or 5; for n = 7 , suitable values for k are 3, 4, and, to much lesser extent, 5; for n = 6 one should choose k = 3 or 4, and so on.
Set the selection of the first k terms of the original polynomial equal to zero. This will be an equation of degree n having zero as root of order n − k + 1 . Find the remaining k − 1 roots of this equation.
Set the remaining terms of the original polynomial equal to zero. This will be an equation of degree 4 or less if k is chosen correctly. Find the roots of this equation.
This step is not necessary, but is very helpful: sketch the graphs of both parts of the split (2) and mark the roots of the two polynomials determined in steps 2 and 3.
Roots of the original equation can exist only in those subintervals (formed by the roots obtained in steps 2 and 3), where both sides of the split (2) have the same signs. These subintervals must be identified. The number of crossing points of the two parts of the split (2) in a subinterval in which they have the same sign can be determined modulo an even number. As with the split (2), there is a limit on that even number – the left-hand side of the split (2) can have up to four stationary points, while the right-hand side can have up to three (if k is chosen appropriately). Thus, there is a limit on the sign changes of the derivatives of the two parts of the split (2) and hence on the number of crossing points in such a subinterval. Another fact must also be taken into consideration: the polynomial on the left-hand side of the split (2) has higher degree than the polynomial on the right-hand side and tends to ± ∞ faster.

(II) Equations of degree 10 or more

For equations of higher degrees, recursive application is needed.

For the fastest application of the method, select k = 5 consecutive terms from the polynomial, starting with the term with the highest power of x, say n ( n ≥ 10 ).
Set this selection equal to zero. This will be an equation of degree n, having zero as root of order n − 4 . Find the remaining four roots of this equation.
Set the remaining terms of the original polynomial equal to zero. If 10 ≤ n ≤ 14 , this will result in an equation of degree between 5 and 9. Apply (I) to this equation and return with its root bounds. Roots of the equation of degree between 10 and 14 can exist only in those subintervals where both the polynomial from step 2 and the one from this step have the same signs. The bounds on these subintervals must be identified. To find the bounds of the roots of the equation of degree between 10 and 14, the rest of step 5 of (I) applies to these two polynomials (with the right-hand side of the split (2) having up to eight stationary points this time).
If the polynomial determined in step 3 is of degree greater than 9 (i.e. if n > 14 ), apply the preceding steps until step 3 applies, i.e. an equation of degree between 5 and 9 is reached (bottom “rung”). There will be a different split of the type (2) each time the above is done.
Use the obtained bounds to determine bounds of the roots for the equation of degree between 15 and 19 (third last rung), preceding the equation of degree between 10 and 14, by following the same procedure.
Repeat until the original equation is reached.

Note that one can avoid solving quartic (and even cubic) equations. For example, the split (2) could be applied to equations of degree n = 5 or less by splitting them with k = 3 or less in such a way that only quadratic equations will need to be solved. Then, for equations of degree 6 or more, the split (2) should be applied recursively in an analogous manner.

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Received: 2020-04-03

Revised: 2020-08-16

Accepted: 2020-09-10

Published Online: 2020-12-01

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2020-0079

Keywords for this article

roots of equations; isolation intervals; Descartes rule

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