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.	$(3x^2)/(x)=3x$	$\begin{array}{cccc}x-4& 3x^2& 3x& -4\\ \text{HLINE TBD}& 3x^2& 3x& \\ 3x(x-4)& 3x^2& -12x& \\ & & 15x& -4\\ 15(x-4)& & 15x& -60\\ & & & 56\end{array}$
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:	$\begin{array}{ccc}Q(x)=3x+15& & R=56\end{array}$

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 $(3x+1)$ .