Polynomial division (HL)

Polynomial division is essential for understanding (and integrating) rational functions, applying remainder theorem, and solving a cubic. Though performing a polynomial division is not required, as it will only appear as one of two or more possible methods to a problem or a step.

Contents

Overview

p(x)=d(x)q(x)+r(x)p(x) = d(x)q(x) + r(x)
p(x)d(x)=q(x)+r(x)d(x)\frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)}

p,d,q,rp, d, q, r are polynomials. pp is divided by divisor dd to get some quotient qq and remainder rr. The degree of rr must be smaller than dd. Polynomial division is only useful when degree of pp is at least as large as that of dd. When this is not the case, partial fractions may be useful instead.

Example: Perform the division 2x3+4x2−5x2+3\displaystyle \frac{2x^3 + 4x^2 - 5}{x^2 + 3}.

Long division is presented here. You may wish to use synthetic division  if you prefer.

2x+4x2+3)2x3+4x2+0x−5−2x3+0x2+6x4x2−6x−5−4x2+0x+12−6x−17\begin{matrix} & 2x & +4 & & \\ \hline x^2 + 3 \left)\right.& 2x^3 &+4x^2 &+0x &-5 \\ -& 2x^3 &+0x^2 &+6x & \\ \hline & & 4x^2 &- 6x & -5 \\ -& & 4x^2 &+ 0x & +12 \\ \hline & & & -6x &-17 \end{matrix}
2x3+4x2−5x2+3=2x+4+−6x−17x2+3■\frac{2x^3 + 4x^2 - 5}{x^2 + 3} = 2x + 4 + \frac{-6x - 17}{x^2 + 3} \qed

As indicated in the above example, you need to explicitly add missing terms to the dividend when some coefficients are zero.

Cases

If you are learning polynomial division, be sure to cover all following cases

  • dividend up to degree 33 and divisor up to degree 22
  • with and without remainders
  • with some coefficients as 00 or omitted
  • divisor leading coefficient is not 11
  • divisor leading coefficient is greater than that of the dividend