Question

If a polynomial p(x) divided by x-2 then remainder is one and when it is divided by x+1 the remainder is -2 find the remainder when the given polynomial divided by (x-2)(x+1)

Answer 1

Note that $(x-2)(x+1)=x^2-x-2$ is quadratic, so if we divide another polynomial by it then the remainder will be 0, a non-zero constant or a linear polynomial. Any remainder of greater degree could be divided further.

Suppose $p(x)=(x-2)(x+1)g(x)+kx+b$

Then

$$\dfrac{p(x)}{x-2}=(x+1)g(x)+\dfrac{kx+b}{x-2}$$

$$=(x+1)g(x)+k+\dfrac{2k+b}{x-2}$$

If a polynomial $p(x)$ is divided by x-2 then remainder is one.

Hence $2k+b=1$

$$\dfrac{p(x)}{x+1}=(x-2)g(x)+\dfrac{kx+b}{x+1}$$

$$=(x-2)g(x)+k+\dfrac{-k+b}{x+1}$$

If a polynomial $p(x)$ is divided by x+1 then remainder is $-2.$

Hence $-k+b=-2.$

$$\begin{matrix} 2k+b=1 \\ -k+b=-2 \end{matrix}$$

$$\begin{matrix} 3k=3 \\ b=k-2 \end{matrix}$$

$$\begin{matrix} k=1 \\ b=-1 \end{matrix}$$

So the remainder when the given polynomial divided by (x-2)(x+1) is $x-1.$