A new approach for solving polynomial equations is presented in this study. Two techniques for solving quartic equations are described that are based on a new method which was recently developed for solving cubic equations. Higher order polynomial equations are solved by using a new and efficient algorithmic technique. The proposed methods rely on initially identifying the vicinities of the roots and do not require the use of complicated formulas, roots of complex numbers, or application of graphs. It is proposed that under the stated conditions, the methods presented provide efficient techniques to find the roots of polynomial equations.
Various methods have been developed to solve polynomial equations (e.g. [
Many techniques have been proposed for solving quartic equations (e.g. [
A quartic equation can be written as:
f ( x ) = x 4 + α 1 x 3 + α 2 x 2 + α 3 x + α 4 = 0 (1)
If x = z − ( α 1 / 4 ) , the above equation can be expressed in terms of z as a depressed quartic equation:
f ( z ) = z 4 + a z 2 + b z + c (2)
where:
a = α 2 − 3 ( α 1 2 / 8 )
b = α 3 − ( α 1 α 2 / 2 ) + ( α 1 3 / 8 )
c = α 4 − ( α 1 α 3 / 4 ) + ( α 1 2 α 2 / 16 ) − 3 ( α 1 4 / 256 )
Equation (2) can be factored into two quadratic equations as found originally by Descartes [
If R1 and R2 are two roots of this quartic equation:
f ( z ) = ( z − R 1 ) ( z − R 2 ) ( z 2 + p z + q ) = [ z 2 − ( R 1 + R 2 ) z + R 1 R 2 ] [ z 2 + p z + q ]
If R 1 + R 2 = γ and R 1 R 2 = λ , comparing this equation with Equation (2) yields:
p = R 1 + R 2 = γ
and:
a = q + λ − γ 2 (3)
b = γ ( λ − q ) (4)
c = q λ (5)
Rearranging Equations (3) and (4) and taking the resulting equations to power 2:
( a + γ 2 ) 2 = ( λ + q ) 2 (6)
( b / γ ) 2 = ( λ − q ) 2 (7)
Subtracting Equation (7) from (6) and substituting qλ by c from Equation (5):
( a + γ 2 ) 2 − ( b 2 / γ 2 ) = 4 c (8)
Multiplying this equation by γ2 yields:
γ 6 + 2 a γ 4 + ( a 2 − 4 c ) γ 2 − b 2 = 0 (9)
If γ 2 = t , Equation (9) can be re-written as:
f ( t ) = t 3 + δ 1 t 2 + δ 2 t + δ 3 = 0 (10)
where:
δ 1 = 2 a
δ 2 = a 2 − 4 c
δ 3 = − b 2
The cubic Equation (10) has at least one real root t1. That root can be found by using the method described in [
a + t 1 = q + λ
and q = c/λ from Equation (5). Therefore:
a + t 1 = λ + ( c / λ ) , hence:
λ 2 − ( a + t 1 ) λ + c = 0 (11)
which can be solved for λ.
After having found t1 and finding λ = R 1 R 2 :
R 1 + R 2 = t 1
R 1 R 2 = λ
These equations can be solved for R1 and R2 which are two roots of Equation (2). The remaining roots of that equation, R3, and R4, can be found by solving:
z 2 + t 1 0.5 z + ( c / λ ) = 0 (12)
Then the roots of Equation (1) can be found by subtracting α1/4 from the values of the roots of f(z).
This method can be used to solve quartic equations with real roots. The minimum and maximum points of the quartic function can be found by differentiating f(x) in Equation (1) with respect to x and setting the resulting equation to zero:
g ( x ) = [ d f ( x ) / d x ] / 4 = x 3 + 0.75 α 1 x 2 + 0.5 α 2 x + 0.25 α 3 = 0 (13)
The roots of this cubic equation can be found by using the method described in [
1) Determining the root types
Theorem 1. The quartic equation with real coefficients f(x) has real roots if α 4 < 0 .
Proof: If α 4 < 0 , then at x = 0 , f ( x ) = α 4 < 0
lim x → ± ∞ f ( x ) = x 4 = + ∞
f(x) is a continuous function of x on (−∞, +∞). Therefore, for an 0 < x 1 < ∞ , f ( x 1 ) = 0 and x 1 ∈ R where R denotes the set of real numbers.
Theorem 2. If f ( x ) > 0 at its minimum points, f(x) has no real root.
Proof: f ( x ) > 0 at its minimum points, and:
lim x → ± ∞ f ( x ) = x 4 = + ∞
There fore, f ( x ) > 0 for all x ∈ R .
Theorem 3. If f(x) has one minimum point, xmin, and f ( x min ) < 0 , f(x) has two real roots.
Proof: x min ∈ R , f ( x min ) < 0 and:
lim x → ± ∞ f ( x ) = x 4 = + ∞
Therefore for a continuous function, f(x), there is an x 1 ∈ R , and − ∞ < x 1 < x min , so that f ( x 1 ) = 0 , and there is an x 2 ∈ R , and x min < x 2 < + ∞ , so that f ( x 2 ) = 0 .
Theorem 4. If x1min, x2min, and x3max are three real roots of g(x) as denoted above, where x 1 min < x 3 max < x 2 min , and f ( x ) < 0 at x1min, and at x2min, and f ( x ) > 0 at x3max, all the roots of f(x) are real.
Proof:
f ( x 1 min ) < 0 , f ( x 3 max ) > 0 , and f ( x 2 min ) < 0 and
lim x → ± ∞ f ( x ) = x 4 = + ∞
f(x) is a continuous function of x on (−∞, +∞).
Therefore,
there is an x 1 ∈ R , and − ∞ < x 1 < x 1 min , so that f ( x 1 ) = 0 ,
there is an x 2 ∈ R , and x 1 min < x 2 < x 3 max , so that f ( x 2 ) = 0 ,
there is an x 3 ∈ R , and x 3 max < x 3 < x 2 min , so that f ( x 3 ) = 0 ,
there is an x 4 ∈ R , and x 2 min < x 4 < + ∞ , so that f ( x 4 ) = 0 .
After determining the type of the roots of f(x), the general procedure of finding the real roots can be followed as explained below.
2) Procedure
a) If f(x) has two real roots:
i) If α4 > 0, one real root of f(x) is between a minimum point and x = 0. The point x = 0 and the minimum point can be chosen as an upper and a lower limit for the real root r1 respectively. Successive iterations can be used to find r1 in that vicinity. f(x) is then divided by (x − r1) and the procedure described in [
ii) If α4 < 0, a line joining a minimum point to x = 0 can be formulated and the intersection of that line with the horizontal axis may be chosen as an upper limit of the root if f(x) is positive at that point. That point and the minimum point can be chosen as an upper and a lower limit of the real root r1 respectively. Successive iterations can be used to find r1 in that vicinity. f(x) will then be divided by (x − r1) and the procedure described in [
b) If f(x) has four real roots:
iii) The maximum and the first minimum points can be chosen as an upper and a lower limit of one real root respectively. Successive iterations can be used to find that root. Similarly, the maximum and the second minimum point can be used as an upper and a lower limit for another real root. Successive iterations can be used to find that root. Then f(x) can be divided by [ x 2 − ( r 1 + r 2 ) x + r 1 r 2 ] where r1 and r2 are the two determined real roots, and the remaining roots of the resulting quadratic equation can be found after the factorization.
An nth order single variable polynomial equation with real coefficients can be written as:
f ( x ) = x n + α 1 x n − 1 + α 2 x n − 2 + α 3 x n − 3 + ⋯ + α n = 0 (14)
where α 1 , ⋯ , α n are real numbers.
If the order of the equation is odd, it is bound to have one or more real roots. If n is even and α n < 0 , the polynomial has real roots. If n is even and α n > 0 , the polynomial may or may not have real roots.
The value of f(x) can be found at 0, −1, and +1 to determine whether it has any roots at these points. If such roots are found, they can be separated by factorization. The value of f(x) can also be found in the close vicinity of −1 and +1 (e.g. at −1.1, −0.9, 0, +0.9, and +1.1, etc.) to see if there are sign changes in f(x). If there are, then successive iterations can be used to find a real root. Then by factorization that root can be separated, and focus can be made on roots whose magnitudes are considerably less than or greater than 1.
Then for a root r, two conditions can hold:
1) If | r | < 1 1
For a high order polynomial, if a root’s magnitude is significantly less than 1, as the order is increased, the polynomial will function as a subset of a Levi-Civita field [
f ( x ) ≈ α n − 2 r 2 + α n − 1 r + α n (15)
Setting this equation to zero, its roots are:
r = [ − α n − 1 ± s 1 ] / 2 α n − 2 (16)
where s 1 = [ k 1 ] 0.5 and k 1 = ( α n − 1 2 − 4 α n − 2 α n ) , and in order for r to be real, k 1 ≥ 0 .
As a rough estimate, if the second order term in Equation (15) can be ignored, r ≈ − α n / α n − 1 .
Then it can be concluded that | α n / α n − 1 | should be less than 1.
If this condition holds, and k 1 ≥ 0 , the values of f(x) at the limits given by Equation (16) and also at x = − ( α n / α n − 1 ) can be checked. If at two of the points, f(x) has opposite signs, a real root between the points can be found by successive iterations. If k 1 < 0 , but | α n / α n − 1 | < 1 , the value of the first derivative of f(x), ( f ′ ( x ) ) can be found at ( − α n / α n − 1 ), and a line formulated at this point with the slope at the point given by f ′ ( x ) . At the intersection of that line with the x-axis, f(x) can be checked. If f(x) at this point and f(x) at ( − α n / α n − 1 ) have opposite signs, a real root between the two points can be found.
2) If | r | > 1
For a high order polynomial (n ≥ 5):
f ( x ) ≈ r n + α 1 r n − 1 + α 2 r n − 2 = r n − 2 ( r 2 + α 1 r + α 2 ) (17)
The non-zero roots of this equation are at:
r = [ − α 1 ± s 2 ] / 2 (18)
where s 2 = [ k 2 ] 0.5 and k 2 = α 1 2 − 4 α 2 , and in order for r to be real, k 2 ≥ 0 .
If the lowest order of n in Equation (17) is ignored, as a rough approximation, r ≈ − α 1 which means that under the conditions, | α 1 | should be greater than 1. If this condition holds and k 2 ≥ 0 , f(x) can be found at the limits given by Equation (18) and also at x = − α 1 . If f(x) at two of the points has opposite signs, a real root between the points can be found by successive iterations. If k 2 < 0 , but | α 1 | > 1 , f ′ ( x ) at x = − α 1 can be found and a line formulated at x = − α 1 with a slope given by f ′ ( x ) at that point. If the intersection of that line with the x-axis and the point x = − α 1 produce f(x) with opposite signs, a real root between the points can be found by successive iterations.
ProcedureThe procedure of finding the real roots of an nth order single variable polynomial with real coefficients and n ≥ 5 can be summarized as follows:
· Consider:
· f ( x ) = x n + α 1 x n − 1 + α 2 x n − 2 + α 3 x n − 3 + ⋯ + α n = 0 .
· Determine f(x) at 0, −1, +1. If there is a root, separate it by factorization.
· Determine f(x) in the vicinity of | r | = 1 (e.g. −1.1, −0.9, +0.9, and +1.1). If there are sign changes in that region, use successive iterations between two points giving opposite signs for f(x) to find a real root and separate that root by factorization.
Is | α n / α n − 1 | < 1 ?
· If so, calculate k 1 = ( α n − 1 2 − 4 α n − 2 α n ) .
· Is k 1 ≥ 0 ?
· If k 1 > 0 , find f(x) at:
· r = [ − α n − 1 ± s 1 ] / 2 α n − 2 .
· where s 1 = [ k 1 ] 0.5 ,
· Calculate f(x) at x = ( − α n / α n − 1 ) .
· If at two of these points, f(x) has opposite signs, find a real root between those points by successive iterations (if k 1 = 0 , use the points − α n − 1 / 2 α n − 2 , and − α n / α n − 1 ).
· If k 1 < 0 , find f ′ ( x ) at ( − α n / α n − 1 ) and call it m1. Then find:
· x 1 = ( − α n / α n − 1 ) .
· f ( x 1 ) = y 1 .
· x 2 = x 1 − ( y 1 / m 1 ) .
· Find f(x2). If f(x1) and f(x2) have opposite signs, find a real root between the points by successive iterations and separate it by factorization.
Is | α 1 | > 1 ?
· If so, calculate k 2 = α 1 2 − 4 α 2 .
· Is k 2 ≥ 0 ?
· If k 2 > 0 , calculate f(x) at r = [ − α 1 ± s 2 ] / 2 .
· where s 2 = [ k 2 ] 0.5 ,
· Calculate f(x) at x = − α 1 .
· If f(x) at two of these points has opposite signs, find a real root between those points by successive iterations and separate it by factorization (if k 2 = 0 , use −α1 and −α1/2 as two points).
· If k 2 < 0 ,
· Find f ′ ( x ) at x = − α 1 and call it m2.
· x 3 = − α 1 .
· f ( x 3 ) = y 3 .
· x 4 = x 3 − ( y 3 / m 2 ) .
· Calculate f(x4). If f(x3) and f(x4) have opposite signs, find a real root between x3 and x4 by successive iterations and separate it by factorization.
If | α n / α n − 1 | > 1 and | α 1 | < 1 , focus should be made on finding roots in the vicinity of | r | = 1 2.
When after factorizations the order of the polynomial is reduced to 4, use the procedure described in the preceding section to find the remaining roots.
Example 1
Assume:
f ( x ) = x 5 − 8 x 4 − 21 x 3 + 8 x 2 − 4 x + 240 = 0
Therefore, α 1 = − 8 , α 2 = − 21 , α 3 = 8 , α 4 = − 4 , α 5 = 240 .
It is quickly determined that the polynomial does not have a root at −1, 0, and +1.
| α n / α n − 1 | = 240 / 4 = 60 > 1
| α 1 | = 8 > 1
k 2 = α 1 2 − 4 α 2 = 8 2 − 4 × ( − 21 ) = 148 > 0
s 2 = [ k 2 ] 0.5 = 12.16
Limits of a root = [−α1 ± s2]/2 = −2.08, 10.08
f ( − 2.08 ) = 283.23
f ( 10.08 ) = 978.12
f ( − α 1 ) = f ( 8 ) = − 10032
Starting successive iterations with the points at 8 and 10.08 yields r = 10 after 7 iterations.
By factorization:
f 1 ( x ) = f ( x ) / ( x − 10 ) = x 4 + 2 x 3 − x 2 − 2 x − 24
To find the roots of the quartic function f1(x), one of the methods of solving quartic equations that were described above can be used. If method #1 is used:
x = z − ( 2 / 4 ) = z − 0.5
and f 1 ( z ) = z 4 − 2.5 z 2 − 23.4375 .
The roots of f1(z) are found at z = ±2.5 and the roots of f1(x) are found as:
r 2 = − 2.5 − 0.5 = − 3
r 3 = 2.5 − 0.5 = 2
Then by factorization:
f 1 ( x ) / ( x − 2 ) ( x + 3 ) = x 2 + x + 4
Setting: f 2 ( x ) = x 2 + x + 4 = 0 , yields:
r 4 , r 5 = − 0.5 ± 1.936 i
Therefore, the roots of the 5th order polynomial f(x) are found at:
−3, 2, 10, and (−0.5 ± 1.936i).
Example 2
Assume:
f ( x ) = x 6 − 2.5 x 5 − 24 x 4 + 8.5 x 3 + 38 x 2 + 126 x − 72
Therefore: α 1 = − 2.5 , α 2 = − 24 , α 3 = 8.5 , α 4 = 38 , α 5 = 126 , α 6 = − 72 .
It is determined that f(x) has no root at −1, 0, and 1.
Following the general procedure as described above:
| α n / α n − 1 | = 72 / 126 = 0.57 < 1
k 1 = ( α n − 1 2 − 4 α n − 2 α n ) = ( 126 ) 2 − 4 × 38 × ( − 72 ) = 26820
s 1 = [ 26820 ] 0.5 = 163.768
The limits for a root are calculated as:
[ − α n − 1 ± s 1 ] / 2 α n − 2 = [ − 126 ± 163.768 ] / 2 × 38 = − 3.81 , 0.497
and:
f ( − 3.81 ) = − 461.9
f ( 0.497 ) = − 0.473
f ( 0.57 ) = 11.09
A root of f(x) is found between 0.497 and 0.57 at r1 = 0.5 after 4 iterations.
Following the procedure to find the other roots:
| α 1 | = 2.5 > 1
k 2 = α 1 2 − 4 α 2 = ( 2.5 ) 2 − 4 × ( − 24 ) = 102.25
s 2 = [ k 2 ] 0.5 = 10.11
Limits of a root = [−α1 ± s2]/2 = −3.8, 6.3
f ( 2.5 ) = − 324.18
f ( − 3.8 ) = − 481
f ( 6.3 ) = 4261
The iterations can start with the points at 2.5 and 6.3. After 8 iterations, a root is found at r2 = 6.
By factorization:
f ( x ) / ( x − 0.5 ) ( x − 6 ) = x 4 + 4 x 3 − x 2 − 10 x − 24
The remaining roots are the roots of the quartic function:
f 1 ( x ) = x 4 + 4 x 3 − x 2 − 10 x − 24
Following the procedure of method #2 for solving quartic equations as described above, the roots of this quartic function are found at −4, 2, and (−1 ± 1.414i). Therefore the roots of the 6th order polynomial f(x) are found at: −4, 0.5, 2, 6 and (−1 ± 1.414i).
Example 3
Assume:
f ( x ) = x 6 + 0.8 x 5 + 2.1 x 4 − 1.5 x 3 + 3.1 x 2 + 4 x − 5.1
Therefore, α 1 = 0.8 , α 2 = 2.1 , α 3 = − 1.5 , α 4 = 3.1 , α 5 = 4 , α 6 = − 5.1 .
It can be seen that | α 1 | < 1 , and | α 6 / α 5 | = 1.275 > 1 . Therefore, focus should be made on finding roots with magnitudes in the vicinity of 1. It can be determined that:
f(0) = −5.1, f(1) = 4.4, f(−1) = −2.2. Searching for a root between 0 and 1 results in finding.
r = 0.746 after 11 iterations. Dividing f(x) by (x − 0.746) yields (with three decimal place accuracy):
f ( x ) = ( x − 0.746 ) f 1 (x)
where:
f 1 ( x ) = x 5 + 1.546 x 4 + 3.253 x 3 + 0.927 x 2 + 3.791 x + 6.828
It can be seen that for f1(x), | α 1 | = 1.546 > 1 . Following the above procedure:
k 2 = − 10.62
f ′ 1 ( x ) = 5 x 4 + 6.184 x 3 + 9.759 x 2 + 1.854 x + 3.791
f ′ 1 ( − 1.546 ) = 29.96 = m 2
x 3 = − 1.546 , and f 1 ( x 3 ) = y 3 = − 8.837 , x 4 = − 1.546 + ( 8.837 / 29.96 ) = − 1.25 , and f 1 ( x 4 ) = − 2 , and f 1 ( − 1 ) = 1.257 .
A real root of r = −1.108 can be found between −1, and −1.25 with three decimal place accuracy after 10 iterations. Therefore f(x) has two real roots so far with magnitudes in the vicinity of | r | = 1 . By factorizing f1(x):
f 1 ( x ) = ( x + 1.108 ) f 2 (x)
where:
f 2 ( x ) = x 4 + 0.438 x 3 + 2.768 x 2 − 2.14 x + 6.162
To find the roots of f2(x), the procedures described above for quartic equations can be used. By using method #2, the roots of the derivative function f ′ 2 ( x ) need to be found. The real root of f ′ 2 ( x ) is found after 7 iterations at x = 0.333 by using the method in [
Following the procedure of method #1 for quartic equations to find the roots of f2(x):
x = z − 0.438 / 4 = z − 0.1095
| Condition to Check | Result |
|---|---|
| If g(x) has one real root | f(x) may have one minimum at that point |
| If g(x) has three real roots | f(x) has two minimums and one maximum point |
| If α4 < 0 | f(x) has real roots |
| If α4 = 0 | f(x) has a real root at x = 0 |
| If α4 > 0 | f(x) may or may not have real roots |
| If g(x) has one real root repeated three times, and at that same point f(x) is also zero | that root is a real root of f(x) repeated four times |
| If g(x) has one real root and at that same point f(x) is also zero | that root is a repeated real root of f(x) |
| If f(x) is not zero or negative at any minimum point | the quartic equation has no real root |
| If g(x) has one real root and f(x) is zero or negative at the minimum point | f(x) has two real roots (or one repeated real root) and two complex conjugate roots |
| If g(x) has three real roots, and f(x) is negative at two minimum points and is not negative at the maximum point | all the roots of the quartic equation are real |
Therefore:
f 2 ( z ) = z 4 + a z 2 + b z + c
where:
a = 2.696 , b = − 2.7356 , c = 6.429
If t = γ 2 = ( R 1 + R 2 ) 2 , where R1 and R2 are two roots of f2(z), the following cubic equation will result:
f ( t ) = t 3 + 5.392 t 2 − 18.447 t − 7.4835
A real root of f(t) is found by using the procedure in [
Therefore:
t 1 = ± 1.6266
For R 1 + R 2 = 1.6266 .
If R 1 R 2 = λ , by using Equation (11):
λ 2 − 5.342 λ + 6.429 = 0
The roots of this 2nd order equation are at λ = 1.831 and λ = 3.51.
Then:
R 1 + R 2 = 1.6266 and R 1 R 2 = 1.831 yield:
R 1 , R 2 = 0.8133 ± 1.08145 i
Then by using Equation (12):
z 2 + 1.6266 z + 3.51 = 0
R3 and R4 are the roots of this equation as:
R 3 , R 4 = − 0.8133 ± 1.68776 i
The roots of f2(x) can be found by subtracting 0.1095 from the roots of f2(z) at 0.7038 ± 1.08145i, and at −0.9228 ± 1.68776i. Therefore the roots of f(x) are found at: −1.108, 0.746, −0.9228 ± 1.68776i, and 0.7038 ± 1.08145i.
Two methods for solving quartic equations that use a recently presented technique for solving cubic equations [
formulas, roots of complex numbers, or graphs. The examples presented show how the methods are applied and their efficiencies.
A number of methods have been presented in the literature and are used in practice to find the roots of polynomial equations with real coefficients. A new technique was recently presented to find the roots of cubic functions [
The next phase of this research can focus on further investigation of the efficiency of the presented method for solving high order polynomial equations when some of the coefficients are much larger/smaller than other coefficients.
The author declares no conflicts of interest regarding the publication of this paper.
Tehrani, F.T. (2020) Solution to Polynomial Equations, a New Approach. Applied Mathematics, 11, 53-66. https://doi.org/10.4236/am.2020.112006