Question

Let $f\left(x\right)=x^5+a{x}^3+bx.$ The remainder when $f\left(x\right)$ is divided by $x+1$ is $-3.$ Then the remainder when $f\left(x\right)$ is divided by $x^2-1,$ is

• A. $-3x$
• B. $-3$
• C. $3$
• D. $3x$

Answer 1

The correct option is D $3x$

$f\left(x\right)=x^5+a{x}^3+bx$
When $f\left(x\right)$ is divided by $x+1,$ the remainder is $-3$, so
$f(-1)=-3$
$\Rightarrow -1-a-b=-3\Rightarrow a+b=2\cdots \left(1\right)$

Assuming
$f\left(x\right)=p\left(x\right)\times (x^2-1)+qx+r$
Putting $x=-1,$ we get
$\Rightarrow -3=-q+r\cdots \left(2\right)$
Putting $x=1,$ we get
$\Rightarrow f\left(1\right)=q+r\Rightarrow 1+a+b=q+r$
Using equation $\left(1\right),$ we get
$3=q+r\cdots \left(3\right)$
Solving equations $\left(2\right)$ and $\left(3\right),$ we get
$q=3,r=0$

Therefore, $f\left(x\right)=p\left(x\right)\times (x^2-1)+3x$
Hence, the remainder is $3x$.