Division of Polynomials

Let $A=x^{3}+2x^{2}-1$ and $B=x^{2}-x+1$. Then we can write

$ A=BQ+R $

where $Q=x+3$ and $R=2x-4$ are called the quotient and the remainder,
respectively. You may verify the above equation by expanding and simplifying
the right hand side.

In general, if $A$ and $B$ are two polynomials such that the degree
of $A$ is greater than or equal to the degree of $B$, the process
of finding two polynomial $Q$ and $R$ such that

$ A=BQ+R $

and $R$ is of lower degree than $B$, is called the process of dividing
$A$ by $B$. In this process, $A$ is called the dividend, $B$ the
divisor, $Q$ the quotient, and $R$ the remainder. If $R=0$, we
say $A$ is divisible by $B$.

How the quotient is obtained is best explained in the following example.

Example

Divide $2x^{3}-32x-15$ by $x-3$ and find the quotient and the remainder.

Solution

First we make sure that the dividend and the divisor are written in
descending powers of $x$. Next we divide the first term of the dividend
by the first term of the divisor

$ \frac{2x^{3}}{x}=2x^{2} $

then multiply $2x^{2}$ by the divisor and subtract the result from
the dividend

$\begin{aligned} (2x^{3}-32x-15)-2x^{2}(x-3) & =\cancel{2x^{3}}-32x-15\cancel{-2x^{3}}+6x^{2} & =6x^{2}-32x-15 \end{aligned}$

or using the long division we have

To simplify calculations, we can reverse the signs of the product
of the multiplication and then add it to the dividend; namely

Now we divide $6x^{2}-32x-15$ by $x-3$ and follow the same steps;
that is, we write it in descending power of $x$ and divide its first
term, $6x^{2}$, by the first term of the divisor, $x$: $6x^{2}/x=6x$

and again repeat until the degree of the remainder becomes less than
the degree of the divisor

Therefore

$ 2x^{3}-32x-15=(x-3)(2x^{2}+6x-14)-57. $

Here the quotient is $Q=2x^{2}+6x-14$ and the remainder is $R=-57.$

Example

Divide $x^{5}-4x^{3}+x^{2}+1$ by $x^{2}+2x-1$ and find the quotient
and the remainder

Solution

The quotient and the remainder are $Q=x_{3}-2x^{2}+x-3$ and $R=7x-2$,
respectively.