# High School Mathematics Extensions/Supplementary/Polynomial Division

* Supplementary Chapters
* Content
* Basic Counting
* Polynomial Division
* Partial Fractions
* Summation Sign
* Complex Numbers
* Differentiation
* Problems & Projects
* Problem Set
* Solutions
* Exercise Solutions
* Problem Set Solutions

## Introduction

First of all, we need to incorporate some notions about a much more fundamental concept: factoring.
We can factor numbers,
    $5\times 7=35$
or even expressions involving variables (polynomials),
    $(x-3)(x+7)=x^{2}+4x-21$
Factoring is the process of splitting an expression into a product of simpler expressions. It's a technique we'll be using a lot when working with polynomials.

### Dividing polynomials

There are some cases where dividing polynomials may come as an easy task to do, for instance:
    ${\frac {x^{3}+6x-12}{2x}}$
Distributing,
    ${\frac {x^{3}}{2x}}+{\frac {6x}{2x}}-{\frac {12}{2x}}$
Finally,
    ${\frac {1}{2}}x^{2}+3-{\frac {6}{x}}$
Another trickier example making use of factors:
    ${\frac {2x^{3}+3x^{2}+6x+9}{2x+3}}$
Reordering,
    ${\frac {2x^{3}+6x+3x^{2}+9}{2x+3}}$
Factoring,
    ${\frac {2x(x^{2}+3)+3(x^{2}+3)}{2x+3}}$
One more time,
    ${\frac {(2x+3)(x^{2}+3)}{2x+3}}$
Yielding,
    $x^{2}+3$

## Long division

What about a non-divisible polynomials? Like these ones:
    $(3x^{2}+3x-4)/(x-4)$
Sometimes, we'll have to deal with complex divisions, involving large or non-divisible polynomials. In these cases, we can use the long division method to obtain a quotient, and a remainder:
    $P(x)=Q(x)\times C(x)+R$
In this case:
    $(3x^{2}+3x-4)=Q(x)\times (x-4)+R$
Long division method :
| 1 | We first consider the highest-degree terms from both the dividend and divisor, the result is the first term of our quotient. | $(3x^{2})/(x)=3x$ | ${\begin{array}{r|ccc}x-4&3x^{2}&3x&-4\\\hline \hline &3x^{2}&3x&\\3x(x-4)&3x^{2}&-12x&\\\hline &&15x&-4\\15(x-4)&&15x&-60\\\hline &&&56\\\end{array}}$ |
| 2 | Then we multiply this by our divisor. | $(3x)\times (x-4)=3x^{2}-12x$ | ${\begin{array}{r|ccc}x-4&3x^{2}&3x&-4\\\hline \hline &3x^{2}&3x&\\3x(x-4)&3x^{2}&-12x&\\\hline &&15x&-4\\15(x-4)&&15x&-60\\\hline &&&56\\\end{array}}$ |
| 3 | And subtract the result from our dividend. | $(3x^{2}+3x-4)-(3x^{2}-12x)=15x-4$ | ${\begin{array}{r|ccc}x-4&3x^{2}&3x&-4\\\hline \hline &3x^{2}&3x&\\3x(x-4)&3x^{2}&-12x&\\\hline &&15x&-4\\15(x-4)&&15x&-60\\\hline &&&56\\\end{array}}$ |
| 4 | Now once again with the highest-degree terms of the remaining polynomial, and we got the second term of our quotient. | $(15x)/(x)=15$ | ${\begin{array}{r|ccc}x-4&3x^{2}&3x&-4\\\hline \hline &3x^{2}&3x&\\3x(x-4)&3x^{2}&-12x&\\\hline &&15x&-4\\15(x-4)&&15x&-60\\\hline &&&56\\\end{array}}$ |
| 5 | Multiplying... | $(15)\times (x-4)=15x-60$ | ${\begin{array}{r|ccc}x-4&3x^{2}&3x&-4\\\hline \hline &3x^{2}&3x&\\3x(x-4)&3x^{2}&-12x&\\\hline &&15x&-4\\15(x-4)&&15x&-60\\\hline &&&56\\\end{array}}$ |
| 6 | Subtracting... | $(15x-4)-(15x-60)=56$ | ${\begin{array}{r|ccc}x-4&3x^{2}&3x&-4\\\hline \hline &3x^{2}&3x&\\3x(x-4)&3x^{2}&-12x&\\\hline &&15x&-4\\15(x-4)&&15x&-60\\\hline &&&56\\\end{array}}$ |
| 7 | We are left with a constant term - our remainder: | ${\begin{array}{lcr}Q(x)=3x+15&&R=56\end{array}}$ | ${\begin{array}{r|ccc}x-4&3x^{2}&3x&-4\\\hline \hline &3x^{2}&3x&\\3x(x-4)&3x^{2}&-12x&\\\hline &&15x&-4\\15(x-4)&&15x&-60\\\hline &&&56\\\end{array}}$ |
So finally:
    $(3x^{2}+3x-4)=(3x+15)\times (x-4)+56$