Question

Let $f\left(x\right)$ be a polynomial function. If $f\left(x\right)$ is divided by $x-1,x+1$ and $x+2$, then remainders are $5,3$ and $2$ respectively. When $f\left(x\right)$ is divided by $x^3+2x^2-x-2$, then remainder is

• A. $x-4$
• B. $x+4$
• C. $x-2$
• D. $x+2$

Answer 1

Answer: (B)

$x+4$

We know that when a polynomial $f\left(x\right)$ is divided by $(x-a)$, the remainder is $f\left(a\right)$

Let $f\left(x\right)$ be the polynomial, $q\left(x\right)$ be the quotient and $r\left(x\right)$ be the remainder.

Degree of remainder is always less than divisor. So, let $r\left(x\right)=a{x}^2+bx+c$

Given that $f\left(1\right)=5;f(-1)=3;f(-2)=2$

$\Rightarrow x^3+2x^2-x-2=(x-1)(x+1)(x+2)$

$\Rightarrow f\left(x\right)=q\left(x\right)\times (x-1)(x+1)(x+2)+r\left(x\right)$

$\Rightarrow f\left(1\right)=q\left(1\right)\times 0+r\left(1\right)$

$\Rightarrow 5=r\left(1\right)$ ..$\left(1\right)$

$\Rightarrow f(-1)=q(-1)\times 0+r(-1)$

$\Rightarrow 3=r(-1)$ ..$\left(2\right)$

$\Rightarrow f(-2)=q(-2)\times 0+r(-2)$

$\Rightarrow 2=r(-2)$ ..$\left(3\right)$

Substituting these values from $\left(1\right),\left(2\right),\left(3\right)$ in $r\left(x\right)$ we get

$\Rightarrow r\left(x\right)=a{x}^2+bx+c$

$\Rightarrow r\left(1\right)=a+b+c=5$

$\Rightarrow r(-1)=a-b+c=3$

$\Rightarrow r(-2)=4a-2b+c=2$

Solving for $a,b,c$ we get

$\Rightarrow a=0;b=1;c=4$

Therefore, $r\left(x\right)=x+4$