On the Determination of the Number of Positive and Negative Polynomial Zeros and Their Isolation

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 and also isolate quite successfully the zeros of the polynomial. 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 no more than four. For polynomials of higher degree, either of the two variations of the method should be applied recursively. Full classification of the roots of the cubic equation, together with their isolation intervals, is presented. Numerous examples are given.


Introduction
An algebraic equation with real co-efficients cannot have more positive real roots than the sequence of its co-efficients has variations of sign is the statement of Descartes' original rule of signs [1] from 1637. 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 co-efficients, or is equal to the number of variations of signs in the sequence of the co-efficients reduced by an even number. Many extensions of the Descartes rule have been proposed -see [3], where 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 → (ax + 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 [(ax + 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 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 co-efficients and of degree n. The method also allows to find 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 gives 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 that of p n (x)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 p n (x) 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 (almost) 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 co-efficient 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 co-efficient 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 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 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.

The Method
Consider the equation a n x n + a n−1 x n−1 + . . . + a 1 x + a 0 = 0 (1) and write the corresponding polynomial p n (x) as p n (x) = a n x n + a n−1 x n−1 + . . .
where 0 < k < n and: f n (x) = x k (a n x n−k + a n−1 x n−k−1 + . . . 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, divide 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 signs and this also allows to count 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, un 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 of 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 four. 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 co-efficient a 0 will be set equal to 1. It suffices to say that, should the equation p n (x) = 0 has 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: p n (x) = a n x n + a n−1 x n−1 + . . . where: q n (x) = a n x n + 1, 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 co-efficient 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 co-efficients 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 co-efficients a j , j = 1, 2, . . . , n − 1, are as they were given through the equation. Calculate next the discriminant ∆ 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 ∆ n (α) = 0 is an equation in α of degree n − 1. Denote the n − 1 roots of the equation ∆ n (α) = 0 by α 1 , α 2 . . . , α n−1 . Then, for the real roots of ∆ n (α) = 0, each of the equations 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 ∆ 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 ∆ 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 2m or has just one real root if it is of order 2m + 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 to find Substituting into q n (χ) = r n−1 (χ) yields: −(n − 1)a n χ n − (n − 2)a n−1 χ n−1 − . . . − a 3 χ 3 − a 2 χ 2 + 1 = 0.
The roots χ j of this equation are the same as the double roots of the equations (8). Equation (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 to determine the exact number of positive and negative roots of the equation but to also isolate them. For equation of degree 6 or higher, this variant of the method should be used recursively.

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 ax + 1 = 0, the only root is determined by the intersection of the straight line y = ax + 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.
(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/dx)q 2 (x)| (x=χ,b=β) = (d/dx)r 1 (x)| (x=χ,b=β) . That is, one has the two simultaneous equations aχ 2 + 1 = −βχ and 2aχ = −β, from which it is easily determined that β = −2 √ a and χ = −β/2a = 1/ √ a.
The resulting root χ of the quadratic equation is double. This corresponds to vanishing discriminant ∆ 2 = b 2 − 4a. For −2 √ a < b ≤ 0 (i.e. when the discriminant ∆ 2 is negative), there are no real roots of the quadratic equation (this includes the depressed equation ax 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 ax 2 + bx + 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.
sponding to the depressed equation ax 2 + 1 = 0), and b < 0. The roots of the quadratic equation are real in each of the three subcases. 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) = ax 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 −∞ 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.

Cubic Equations
Consider a cubic equation which does not have a zero root. Without loss of generality, such equation can be written as p 3 (x) = ax 3 + bx 2 + cx + 1 = 0 or as either one of the following two splits: That is: Provided that b = 0, the graph of the quadratic polynomial r 2 (x) = −bx 2 − cx passes through the origin and also through point −c/b from the abscissa. The graph of the cubic polynomial q 3 (x) = ax 3 + 1 passes through point (0, 1) and also through point − 3 1/a from the abscissa. The cubic polynomial ψ 3 (x) = ax 3 + bx 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

The Depressed Cubic Equation
Firstly, 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 αx 3 + cx + 1 = 0 would have a double root, consider the discriminant −α(4c 3 + 27α). This vanishes when α is either 0 or −4c 3 /27. The cubic equation αx 3 + cx + 1 = 0 with α = −4c 3 /27 has root χ 0 = 3/c and a double root χ = −3/(2c). One can then immediately determine the number of positive and negative roots of the original equation ax 3 + cx + 1 = 0 and also localise them as follows (see 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 ax 3 + 1 with the abscissa and the origin. Namely, 3 −1/a < x 0 < 0. The other two roots are complex. 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 ax 3 + 1 with the abscissa. Namely, abscissa. Namely, x 0 > 3 −1/a. If, further, a < α, then the equation has no other roots. If a = α, the equation has an additional double root given by χ = −3/2c < 0. Finally, if a > α, then the equation has, additionally, two negative roots: one between 0 and χ = −3/2c, the other -smaller than χ = −3/2c. Again, one only needs to compare a, given through the equation, to α = −4c 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 ax 3 + 1 with the abscissa, that is, 0 < x 0 < 3 −1/a < x 0 . As a numerical example of the above, consider the equation The auxiliary equation has discriminant −α(27α − 256). Clearly, when α = 256/27, equation (15) will have a double root χ. To find χ, one needs to solve the equation (11): 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 equation (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 3x 3 + 1 and the abscissa, that is The roots of equation (14) are:

Full Cubic Equation
If one considers next the cubic equation ax 3 + bx 2 + 1 = 0, the situation will not turn out to be qualitatively different from the one of the "full" equation ax 3 + bx 2 + cx + 1 = 0, where all co-efficients are different from zero, and the latter is the equation to be considered next.
The discriminant of the cubic equation is given by Setting ∆ 3 = 0 and interpreting b and c as parameters, one gets a quadratic equation for a with roots These are real, i.e. the discriminant ∆ 3 can be zero, only when c 2 > 3b. Let α 1 denote the bigger root. If, further, c 2 > 4b, then the free term in the quadratic equation ∆ 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 = 4b, in which case the roots (19) are 0 and c 3 /54, and c 2 = 3b, in which case equation ∆ 3 = 0 has a double root c 3 /27. For the case of a general equation of degree 3, equation (11) becomes: The roots of this equation are At point x = χ 1,2 , the curve α 1,2 x 3 + 1 is tangent to the curve −bx 2 − cx 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 localise 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, the analysis for each of the remaining four cases can be easily inferred afterwards.
This is the most complicated case. There are three sub-cases.
In this sub-case, the negative root of −bx 2 − cx is to the left of the point where ax 3 + 1 intersects the abscissa -see Figure 3a.
If one further has c 2 ≥ 4b, 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 −bx 2 −cx, but the given equation Thus, the double root χ of equation α 2 x 3 + 1 = −bx 2 − cx will be given by χ 1 , while the other root (using Viète's formula) When c 2 > 4b, the curve ax 3 + 1 with a = 0 will intersect the curve −bx 2 − cx between −c/(2b) and the origin at point σ which is the bigger root of the quadratic equation Thus, for any a greater than 0, provided that −c/b < − 3 1/a, the curve ax 3 + 1 will intersect the curve −bx 2 − cx, for which c 2 ≥ 4b, once between σ and 0. As a sub-case with −c/b < − 3 1/a 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 is the smaller root of −bx 2 − cx = 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 ≥ 4b, and −c/b < − 3 1/a the following two equations.
Indeed, the roots are −1.938, −1.494, and −0.897 approximately. Finally, when c 2 < 3b in the sub-case of −c/b < − 3 1/a, then the cubic equation will have one negative root between − 3 1/a and 0 and two complex roots. This is illustrated by The model predicts a negative root between − 3 1/a ≈ −0.794 and 0 and two complex roots. Indeed, one has x 1 = −0.739 and x 2,3 = 0.119 ± 0.814 i.
In this sub-case, the negative root of −bx 2 − cx is between the point where ax 3 + 1 intersects the abscissa and the origin. If in this case one also has c 2 > 4b, then the curve ax 3 + 1 with a = 0 will intersect the curve −bx 2 − cx between −c/b and the origin at two points: the roots σ 1,2 of the quadratic equation −bx 2 − cx = 1, i.e. σ 1,2 = [−c/(2b)](1 ± 1 − 4b/c 2 ) < 0 (with σ 2 < σ 1 ). These are on either side of −c/(2b). Then for any a greater than 0, given that −c/b > − 3 1/a, the curve ax 3 + 1 will intersect the curve −bx 2 − cx with c 2 ≥ 4b, 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 − 3 1/a. Therefore, the cubic equation will have 3 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 − 3 1/a. A double root χ cannot exist when c 2 ≥ 4b. Equation with a > 0, b > 0, c > 0, c 2 ≥ 4b, and −c/b > − 3 1/a can be illustrated with the following numerical example: 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 − 3 1/a ≈ −3.107. Next, if one has 3b ≤ c 2 < 4b instead of c 2 ≥ 4b, the situation on Figure 3c applies. In view of the restriction −c/b > − 3 1/a, one can have: either a < α 2 , or a = α 2 , or α 2 < a < (b/c) 3 . In the first case, the cubic equation will have a negative root smaller than − 3 1/a and two complex roots. In the second case, the cubic equation will have a negative root smaller than − 3 1/a and a double negative root ar 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 = 3b, 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, 3b ≤ c 2 < 4b, and −c/b > − 3 1/a are given. The first one is: 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 ≥ 4b, one has c 2 < 3b (see Figure 3b for the case of 3b < c 2 < 4b), then the cubic discriminant ∆ 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 = 3b, then the double root χ is at −c/b = −3/c and α 1 = α 2 = c 3 /27.

The Sub-case of
. It is negative when c 2 < 3b and in this case the cubic equation will have a negative root given by x = −c/b = 3 −1/a together with two complex roots. If the discriminant ∆ 3 is positive (i.e. c 2 > 3b), the cubic equation will have three negative roots given by: Firstly, the equation which is in the category c 2 < 3b 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, equation is with c 2 > 3b and, since a = (b/c) 3 (11). The other root, (c/b)(−1 + 1 − 3b/c 2 ) < −c/b > 0, is associated with the irrelevant α 2 < 0. When a is smaller than α 1 , the biggest root is positive and is between χ and σ 1 = [−c/(2b)](1 + 1 − 4b/c 2 ) which is the bigger root of −bx 2 − cx − 1 = 0. The middle root is also positive and is between σ 2 = [−c/(2b)](1− 1 − 4b/c 2 ) (which is the smaller root of −bx 2 − cx− 1 = 0) and χ. The third root is negative and is smaller than − 3 1/a. If a = α 1 , there is a positive double root χ and a negative root smaller than − 3 1/a. Finally, if a > α 1 , then the cubic discriminant ∆ 3 is negative and the equation has a negative root smaller than − 3 1/a together with two complex roots. When c 2 = 4b, the bigger root in (19) is α 1 = 0. The other one is α 2 = c 3 /54 < 0. Thus the curve ax 3 + 1 with a = α 1 = 0 is tangent to the curve −bx 2 − cx at point −c/(2b) where the maximum of −bx 2 − cx occurs (the maximum in this case is c 2 /4b = 1). As a cannot be zero (one has to have an equation of degree 3), the cubic equation has a negative root smaller than − 3 1/a and two complex roots. If 3b ≤ c 2 < 4b, the roots α 1,2 will have the same signs. The co-efficient in the term linear in a is 2c(9b − 2c 2 ) and in the case of 3b ≤ c 2 < 4b, 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 ax 3 + 1 with a > 0 can intersect in the first quadrant the curve −bx 2 − cx with c 2 ≤ 4b. The cubic discriminant ∆ 3 is non-negative. The roots of the cubic equation are as in the latter case: two complex and one negative and smaller than − 3 1/a. When c 2 < 3b, the discriminant ∆ 3 is negative and no curve ax 3 + 1 with any a could be tangent to the curve −bx 2 − cx. The roots of the cubic equation are, again, two complex and one negative and smaller than − 3 1/a. This case will be illustrated with the following four examples. Consider first the equation The roots ( Both are irrelevant, since the only curves ax 3 + 1 that can be tangent to −bx 2 − cx are those with a < 0 while the considered a is positive. The equation should have two complex roots and a negative root smaller than − 3 1/a = −1. This is the case indeed: x 1,2 ≈ 0.492 ± 0.305 i and x 3 ≈ −2.984. Finally, the equation is with c 2 < 3b. The discriminant ∆ 3 is negative. Thus, there should be a root smaller than − 3 1/a ≈ −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 = Then the roots of the cubic equation are as follows. There is always one negative root between − 3 1/a 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 − 3 1/a and 0, there is a positive double root at χ. If a < α 1 , the roots are: one negative root between − 3 1/a and 0; one positive root between σ 1 and χ; and another positive root greater than χ -see Figure 4b.
The given a = 2 is greater than α 1 = 1, thus the equation must have one negative root between c/b = −1 and − 3 1/a ≈ −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 is another example chosen so that one again has: The given a = 1/2 is now smaller than α 1 = 1, thus the equation must have one negative root between − 3 1/a ≈ −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.

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).

Recursive Application of the Method. An Example with Equation of Degree 7
Consider the equation of seventh degree and see Figure 6.
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 co-efficient 16/3 of x 7 ) will be of degree 6 and not solvable analytically. One has to proceed by applying the method twice. Firstly, re-write the given equation µ(x) = − 17 6 x − 104 17 x 5 + 28 17 The x 5 + 28 17 and split further:  Consider the following example as an equation of degree above 5 and up to and including 9: x 9 + 1 2 x 8 − 7x 7 − 2x 6 + 9x 5 − x 4 − 2x 3 + 13x 2 + 14x − 24 = 0.
Using the split (2), this equation can be written as f (x) − g(x) = 0 with 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. Further, when x → −∞, f (x) → −∞, while when x → ∞, f (x) → ∞. It is essential to note that f (x) can have up to 4 non-zero extremal points. Namely, f ′ (x) = 9x 8 + 4x 7 − 49x 6 − 12x 5 + 45x 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) = 4x 3 + 6x 2 − 26x − 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 sketchedsee 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 2 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. The Lagrange bound provides that all roots are between ± max 1, n−1 i=0 |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 Different Perspective on the Split (5)
Given the equation a n x n + a n−1 x n−1 + a n−2 x n−2 + . . . + a 1 x + a 0 = 0 which is of degree n and thus the coefficient a n cannot be zero, one can instead consider an equation with a n set equal to 1: x n + a n−1 x n−1 + a n−2 x n−2 + . . . + a 1 x + a 0 = 0, which has the same roots. Then the split x n + a 0 = −a n−1 x n−1 − a n−2 x n−2 − . . . − a 1 x (60) 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 one smaller than that of the given equation as n − 1 is the highest power of a 0 in the discriminant ∆ 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 ∆ 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.