# High School Mathematics Extensions/Supplementary/Polynomial Division

## 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,

$x^{2}+3$ 1. Try dividing $35x^{2}+29x+6$ by $2.5x+1$ .

2. Now, can you factor $P(x)=3x^{3}-9x+6$ ?


## 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. $(3x)\times (x-4)=3x^{2}-12x$ 3 And subtract the result from our dividend. $(3x^{2}+3x-4)-(3x^{2}-12x)=15x-4$ 4 Now once again with the highest-degree terms of the remaining polynomial, and we got the second term of our quotient. $(15x)/(x)=15$ 5 Multiplying... $(15)\times (x-4)=15x-60$ 6 Subtracting... $(15x-4)-(15x-60)=56$ 7 We are left with a constant term - our remainder: ${\begin{array}{lcr}Q(x)=3x+15&&R=56\end{array}}$ So finally:

$(3x^{2}+3x-4)=(3x+15)\times (x-4)+56$ 3. Find some $G(x)$ such that $(6x^{2}-13x+7)-G(x)$ is divisible by $(3x+1)$ .