### Theory:

The process described in the previous section is somewhat tedious to divide two polynomials, so we use Remainder theorem.
Remainder theorem:

Let $p\left(x\right)$ be any degree polynomial greater than or equal to one and let $$a$$ be any real number. When a linear polynomial  divides $p\left(x\right)$, then $p\left(a\right)$ will be the Remainder.

Proof:

Let $p\left(x\right)$ be any polynomial that is greater than or equal to degree $$1$$.

Assume that the quotient is $$q(x)$$ when $p\left(x\right)$ is divided by  and, the remainder is .

Applying the division algorithm, Dividend $$=$$ Divisor $$\times$$ Quotient $$+$$ Remainder

......................... Equation (i).

Since the degree of  is $$1$$ and the degree of  is less than the degree of , the degree of $\phantom{\rule{0.147em}{0ex}}r\left(x\right)=0$.

This means that is a constant, say $\phantom{\rule{0.147em}{0ex}}r$.

So we can rewrite Equation (i) as   .................... Equation (ii)

In particular, if , then Equation (ii) becomes  .

Thus, $$p(a) = r$$.  Therefore, $$p(a)$$ is the remainder.
Example:
Find the remainder when  is divided by .

Solution: Zero of  is $$1$$, so as per remainder theorem remainder in this case will be $$p(x) =$$ $$p(1)$$ .

Consider the polynomial $\phantom{\rule{0.147em}{0ex}}p\left(x\right) =\phantom{\rule{0.147em}{0ex}}{x\phantom{\rule{0.147em}{0ex}}}^{4}+\phantom{\rule{0.147em}{0ex}}{x}^{3}\phantom{\rule{0.147em}{0ex}}–\phantom{\rule{0.147em}{0ex}}{2x\phantom{\rule{0.147em}{0ex}}}^{2}+\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}1$.

Put $x=1$.

$p\left(1\right) ={\phantom{\rule{0.147em}{0ex}}\left(1\right)}^{4}\phantom{\rule{0.147em}{0ex}}+{\phantom{\rule{0.147em}{0ex}}\left(1\right)}^{3}\phantom{\rule{0.147em}{0ex}}–\phantom{\rule{0.147em}{0ex}}2{\left(1\right)\phantom{\rule{0.147em}{0ex}}}^{2}$ $$+1$$ $$+1$$.

$\begin{array}{l}p\left(1\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}1+1-2+1+1\\ \\ p\left(1\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}2-2+1+1\\ \\ p\left(1\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}2-2+2\\ \\ p\left(1\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}2\end{array}$

So that, the remainder of the polynomial ${\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}}^{4}+\phantom{\rule{0.147em}{0ex}}{x\phantom{\rule{0.147em}{0ex}}}^{3}–\phantom{\rule{0.147em}{0ex}}{2x}^{2}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}x\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}1$ is $$2$$.