High School Mathematics Extensions/Supplementary/Polynomial Division

First of all, we need to incorporate some notions about a much more fundamental concept: factoring.

We can factor numbers,

${\displaystyle 5\times 7=35}$

or even expressions involving variables (polynomials),

${\displaystyle (x-3)(x+7)=x^{2}+4x-21}$

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.

There are some cases where dividing polynomials may come as an easy task to do, for instance:

${\displaystyle {\frac {x^{3}+6x-12}{2x}}}$

Distributing,

${\displaystyle {\frac {x^{3}}{2x}}+{\frac {6x}{2x}}-{\frac {12}{2x}}}$

Finally,

${\displaystyle {\frac {1}{2}}x^{2}+3-{\frac {6}{x}}}$

Another trickier example making use of factors:

${\displaystyle {\frac {2x^{3}+3x^{2}+6x+9}{2x+3}}}$

Reordering,

${\displaystyle {\frac {2x^{3}+6x+3x^{2}+9}{2x+3}}}$

Factoring,

${\displaystyle {\frac {2x(x^{2}+3)+3(x^{2}+3)}{2x+3}}}$

One more time,

${\displaystyle {\frac {(2x+3)(x^{2}+3)}{2x+3}}}$

Yielding,

${\displaystyle x^{2}+3}$

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

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

What about a non-divisible polynomials? Like these ones:

${\displaystyle (3x^{2}+3x-4)/(x-4)}$

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:

${\displaystyle P(x)=Q(x)\times C(x)+R}$

In this case:

${\displaystyle (3x^{2}+3x-4)=Q(x)\times (x-4)+R}$

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

So finally:

${\displaystyle (3x^{2}+3x-4)=(3x+15)\times (x-4)+56}$

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