Question

A polynomial in $x$ of degree greater than $3,$ leaves remainders $2,1$ and $-1$ when divided by $(x-1),(x+2)\text{and}(x+1)$ respectively. What will be the remainder when it is divided by $(x-1)(x+2)(x+1).$

• A. $\frac{7}{6}{x}^2-\frac{3}{2}x-\frac{2}{3}$
• B. $\frac{3}{2}x-\frac{2}{3}$
• C. $\frac{7}{6}{x}^2+\frac{3}{2}x-\frac{2}{3}$
• D. $\frac{3}{2}x+\frac{2}{3}$

Answer 1

The correct option is C $\frac{7}{6}{x}^2+\frac{3}{2}x-\frac{2}{3}$

Given: $2,1$ and $-1$ are remainders when polynomial is divided by $(x-1),(x+2)\text{and}(x+1)$

Now, let the required polynomial be $f\left(x\right)=p\left(x\right)(x-1)(x+2)(x+1)+a_0{x}^2+a_{1}x+a_2$

By remainder theorem, we know $f\left(1\right)=2,f(-2)=1,f(-1)=-1$

$\Rightarrow f\left(1\right)=p\left(1\right)(1-1)(1+2)(1+1)+a_0(1{)}^2+a_1(1)+a_2=2$
$\Rightarrow f\left(1\right)=a_0+a_1+a_2=2\cdots \left(i\right)$

$\Rightarrow f(-2)=p(-2)(-2-1)(-2+2)(-2+1)+a_0(-2{)}^2+a_1(-2)+a_2=1$
$\Rightarrow f(-2)=4a_0-2a_1+a_2=1\cdots \left(ii\right)$

$\Rightarrow f(-1)=p(-1)(-1-1)(-1+2)(-1+1)+a_0(-1{)}^2+a_1(-1)+a_2=-1$
$\Rightarrow f(-1)=a_0-a_1+a_2=-1\cdots \left(iii\right)$

On solving $\left(i\right),\left(ii\right),\left(iii\right)$ we get, $a_0=\frac{7}{6},a_1=\frac{3}{2},a_2=-\frac{2}{3}$

$\therefore f\left(x\right)=p\left(x\right)(x-1)(x+2)(x-1)+\frac{7}{6}{x}^2+\frac{3}{2}x-\frac{2}{3}$

Now if we divide $f\left(x\right)\text{by}(x-1)(x+2)(x+1)$ remainder will be $\frac{7}{6}{x}^2+\frac{3}{2}x-\frac{2}{3}$