Cubic Polynomials Finding Roots Quickly And Easily
By: Greg Oliver
Tags: Quadratic Equation, Math, Mathematics, Mathematics Education, Cubic Polynomial

Cubic Polynomials – A Simpler Approach

An intuitive way to find the 2nd and 3rd roots

While we are all familiar with finding the roots of a quadratic using the Quadratic Equation, it can be complex to find the roots of a higher order polynomial.

We typically need to find one factor by brute force, then divide through to create a quadratic.

I’ve been working on a simpler approach that I’m sharing in this post.

My goal is to make the process a lot simpler by eliminating polynomial long division.

This approach assumes knowledge of algebra and introductory calculus (differentiating polynomials) at the high school level.

I hope it helps you to think in a fresh way about cubic polynomials, and how they can be simpler than first meets the eye!

The method I’ve worked on simply incorporates the known factor into a modified version of the Quadratic Equation, thereby eliminating the division process.

This is the result:

I’ve worked on two methods.

1st Method

The first method is the simpler of the two. It involves writing the polynomial coefficients in factor notation, deriving a Quadratic Equation in terms of the known factor, which can then be solved using a modified version of the standard quadratic formula.

Given y=Ax³+Bx²+Cx+D

Factor notation: y=(x+L)(Ax+M)(x+N)

Expand: y=Ax³+(AN+M+AL)x²+(MN+ALN+LM)x+LMN)

From which for any polynomial: B=(AN+M+AL), C=(LM+LN+MN) and D=LMN

Known factor: Let (𝑥+𝐿) be the known factor. Hence:

y=(x+L)(Ax²+(M+AN)x+MN)

Rewriting the quadratic and substituting B, D and L:

y=Ax²+(M+AN)x+MN

y =Ax²+(B-AL)x +D

Adopting the modified quadratic equation to find the roots, we get the cubic equation;

2nd Method

The second method is constructed on the basis that at the roots of a polynomial, the gradient is given by the product of any one factor, and the gradient of the Quadratic Equation is derived from the other two factors.

Factor notation: y= (x+L)(Ax+M)(x+N)

With the quadratic: y=Ax²+(M+AN)x+MN

Derivative: dy/dx=2Ax+(M+AN)

And cubic:

y=Ax³+(AN+M+AL)x²+(MN+ALN+LM)x+LMN

dy/dx=3Ax²+2(AM+N+AL)x+(MN+ALN+LM)

Now at the roots where (x+L) is the known factor:

dy/dx=3Ax²+2Bx+C=(x+L)(2Ax+(M+AN)

=2Ax²+(M+AN+2AL)x+L(M+AN)

Solving by simultaneous equations:

dy/dx=3Ax²+2Bx+C

dy/dx=2Ax²+(M+AN+AL+AL)x+L(M+AN)

0=Ax²+(2B -M -AN -AL -AL)x+(C -LM -ALN)

Now 𝐵=𝐴𝑁+𝑀+𝐴𝐿, 𝑎𝑛𝑑 D=𝐿𝑀𝑁, substituting these:

𝑦=𝐴𝑥²+(𝐵 -𝐴𝐿)𝑥+𝑀𝑁, (Given C=MN+ALN+LM and LMN = D)

𝑦=𝐴𝑥²+(𝐵 -𝐴𝐿)𝑥+𝐷/𝐿

Therefore:

as before.

A few examples will demonstrate it’s effectiveness:

EXAMPLE 1

y=x³+6x²-x -30

(where A=1, B = 6, C=-1 and D = –30). Given (x+L) = (x+5) is a factor, therefore:

EXAMPLE 2

Let y=4x³+3x²-16x-12

(where A=4, B=3, C=–16 and D=–12). Given (x+L) = (x+2) is a factor, therefore:

=-0.75,+2

EXAMPLE 3

Let y=-x³-3x²+4x+10

Given (𝑥+1.609) is a factor, therefore:

Quartic Polynomials

While cubic polynomials are the primary purpose of this post, I’ve been working on a further modified formula that can be used for higher powers.

However, two known factors are required which may limit its efficiency.

Without going through the lengthy derivation by factor expansions and coefficient reconfigurations etc. (which will be covered in a later post), the quartic formula is as follows

The following example will demonstrate its use;

Let y=x⁴+7x³+5x²-31x-30

(where A=1, B = 7, D=–31 and E=-30)

y=(x+k)(x+l)(Ax+m)(x+n)

Given the known factors: (x+1) and (x+5);

The further modified equation becomes:

Note: the additional term k further adjusts the cubic coefficient of x³ to the quadratic B coefficient of x, in two steps as follows:

B(quad)=(𝐵(𝑞𝑢𝑎𝑟𝑡)−A(𝑘+𝑙)

where for cubics, B(quad)=B(𝐵𝑐𝑢𝑏𝑖𝑐−𝐴𝑙)

E vs. D vs. C is evident.

I’ll post more details soon.

I hope this post has helped you think in a fresh way about cubic polynomials, and an intuitive way to find the 2nd and 3rd roots.