# Find the remainder of $f(x)$ when divided by two binomials

I got stuck on the following two questions:

1. Find the remainder if the polynomial $f(x)$ is divided by $ax+b.$
2. Find the remainder if the polynomial $f(x)$ is divided by $(x-a)(x-b).$

I got as far as rewriting question 1's divisor as $a(x+\frac{b}{a}).$ From there, I didn't know where to proceed.

Whether it's a hint or full solution, I would love some directions on how to proceed.

By the way, the answers are $f(-\frac{b}{a})$ and $(\frac{f(a)-f(b)}{a-b})x+\frac{af(b)-bf(a)}{a-b}.$

• Note that $f(x)=(ax+b)g(x)+r$ [r is constant as the divisor is one-degree]. Put $x=-\frac{b}{a}$ to get $r=f(-\frac{b}{a})$. Try solving the second problem similarly. – JDF Sep 13 '16 at 14:31

In the 1st problem, let $f(x)=(ax+b)g(x)+R$ for some polynomial $g$ and constant $R$.

$=a(x+\frac{b}{a})g(x)+R$

Then the remainder is given by $f(-b/a)=R$

For the 2nd problem,

let $f(x)=(x-a)(x-b)q(x)+Ax+B\quad...(*)$ for some polynomial $q$ and arbitrary constants $A,B$.

We substitute $x=a$ and $x=b$ successively in the identity $(*)$.

Then we have $f(a)-f(b)=Aa-Ab\Rightarrow A=\frac{f(a)-f(b)}{a-b}$

Putting this value in $f(a)$ we get, $f(a)=Aa+B=\frac{af(a)-af(b)}{a-b}+B$

$\Rightarrow B=\frac{af(b)-bf(a)}{a-b}$

So the remainder is given by $Ax+B$

Both problems are elementary applications of the division algorithm and the remainder theorem. Note that the degree of the remainder in both cases can be predetermined - when the divisor is linear, the remainder is constant and when the divisor is quadratic, the remainder is linear.