Question

The remainders left by a polynomial of degree greater than $3$, are $2,1,-1$ when divided by $x-1,x+2$ and $x+1$, then the remainder when polynomial is divided by $(x-1)(x+1)(x+2)$, is:

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

Answer 1

Answer: (D)

$\frac{7x^2}{6}+\frac{3}{2}x-\frac{2}{3}$

Let $f\left(x\right)=(x+1)(x-1)(x-2)Q\left(x\right)+R\left(x\right)$
Here, $f\left(x\right)$ is divided by third degree polynomial, then remainder is quadratic equation
Let $R\left(x\right)=a{x}^2+bx+c$
$\Rightarrow f\left(x\right)=(x+1)(x-1)(x-2)Q\left(x\right)+a{x}^2+bx+c$
Now,Remainder when divided by $(x+1),(x-1),(x-2)$ are $2,1,-1$
$\Rightarrow f\left(1\right)=a+b+c=2$ .... (1)
$f(-2)=4a-2b+c=1$ .... (2)
$f(-1)=a-b+c=-1$ .... (3)
(1) $-$ (3) $\Rightarrow 2b=3\Rightarrow b=\frac{3}{2}$

(2)$-$ (1) $\Rightarrow 3a-3b=-1$
$3a=-1+\frac{9}{2}=\frac{7}{2}$
$a=\frac{7}{6}$
$a+b+c=2$
$\Rightarrow \frac{7}{6}+\frac{3}{2}+c=2$
$\Rightarrow c=2-\frac{7}{6}-\frac{3}{2}=-\frac{2}{3}$
$\Rightarrow R\left(x\right)=\frac{7x^2}{6}+\frac{3x}{2}-\frac{2}{3}$
Hence, option D is correct