Question

If the remainders of the polynomial f(x) whendivided by $x+1,x+2,x-2$ are $6,15,3$ respectively, then the remainder of $f\left(x\right)$ when divided by $(x+1)(x+2)(x-2)$ is

• A. $2x^2-3x+1$
• B. $3x^2-2x+1$
• C. $2x^2-x-3$
• D. $3x^2+2x+1$

Answer 1

The correct option is A $2x^2-3x+1$

let $f\left(x\right)=\theta \left(x\right)(x+1)(x+2)(x-2)+\gamma \left(x\right)$
$\gamma \left(x\right)=a{x}^2+bx+c$
Since order of $D\left(x\right)=3$
So, $f\left(x\right)=\theta \left(x\right)(x+1)(x+2)(x-2)+a{x}^2+bx+c$
$f(-1)=a-b+c=b$
$f(-2)=4a-2b+c=15$
$f\left(2\right)=4a+2b+c=3$
Solving the 3 equations we get
$a=2;b=-3;c=1$
So
$\gamma \left(x\right)=2x^2-3x+1$