Polynomial Division — by Formula. Using a modified Quadratic Equation for…
By: Greg Oliver
Tags: Math, Polynomial, Mathematics, Quadratic Equation, Education

Polynomial Division — by Formula

Using a modified Quadratic Equation for division of higher order polynomials, without long division.

This post presents a method for dividing higher order polynomials, without long division, presenting answers in factor notation.

It uses the variants of the Quadratic Equation for finding roots of higher order polynomials introduced in my post, Cubic Polynomials — A Simpler Approach.

The modified equations I offered bridged the need for polynomial division in finding factors.

This post expands on that methodology to use the modified Quadratic Equation for division of higher order polynomials, without long division. It is most useful for division resulting in quadratic quotients, but with some minor adaptation it can be used for other quotents, as shown in Example 4 below. It doesn’t apply to quotients with remainders.

This post assumes knowledge of algebra at the high school level.

Factor method

Without going through the lengthy derivation by factor expansions and coefficient reconfigurations etc. developed in my earlier post, I’ll start with dividing a quartic polynomial by a quadratic.

In my previous post, the quartic version was presented as follows:

In this, k and l were known factors, resulting in a Quadratic Equation for quartics.

The constant E is divided by k*l since the product of the factors equals the constant.

Division method

In the division application of the formulas outlined here, all factors of the denominator polynomial are reflected in place of k and l. This is because the sum of the factors equals the coefficient of x^n into the numerator polynomial.

Example 1) Quartic Polynomial Division by Quadratic

Divide: y=x⁴+6x³+3x²-12x+4

By: y=x²+2x-1

Insert the coefficients of x^n into the quartic version:

(where A=+1, B=+6, C=+3, D=-12, and E=+4)

k+l=2 and k*l =-1

Roots =-4.83, +0.83; Factors = (x+4.83), (x-0.83)

Solution: y=(x+4.83)(x-0.83)

Example 2) Cubic Polynomial Division by Quadratic

This example is included for completeness, only because the solution is a linear equation, and the quadratic factors are obtainable by the Quadratic Equation:

Divide: y=x³+6x²-x-30

By: y=x²+8x+15

(where A=1, B=6, C=–1, D=-30)

Using the Quadratic Equation, the roots of the quadratic denominator are (x-3), (x-5).

Using either of the factors to the cubic variant we get:

Roots = -3, +2; Factors = (x+3), (x–2)

Solution: y=(x-2)

EXAMPLE 3) Quintic Polynomial Division by Cubic

Note, in this example I have created a cubic to demonstrate the process, because I’m avoiding a remainder.

Divide: y=x⁵-5x⁴+5x³+5x²-6x-1

By: y=x³-4.68x²+4.71x+0.82

Load the quintic/cubic example:

(where A=1, B=-5, C=+5, D=+5, E=-6 and F=-1)

k+l+m=-4.68, and k*l*m=+0.82

Roots= -1.28, +0.95; Factors = (x+1.28), (x-0.95)

Solution: y=(x+1.28), (x-0.95)

EXAMPLE 4) Quintic polynomial divided by quadratic

This example shows another method of dealing with a non-quadratic quotient. It requires one quintic root to be known to proceed.

Divide: y=x⁵-5x⁴+5x³+5x²-6x-1

By: y=x²+0.322x-1.217

In factor notation: y=(x-1.28)(x+0.95)

Given one quintic factor is (x-3.04) and there is no remainder, the known factors of the numerator are: y=(x-1.28)(x+0.95)(x-3.04)

Applying these to the equation we get:

Roots= -0.16, +1.79

Solution: y=(x+0.158)(x-1.79)(x-3.04)

I hope this post has helped you to gain further appreciation of the efficient use of these equations for both factoring and division of polynomials.

If this has interested you, I explore how these ideas can be simplified even further in my post Polynomials — Division by Vision.