Question

The three degree polynomial $f\left(x\right)$ has roots of the equation $3,-3$ and $-k$. Given that the coefficient of $x^3$ is $2$ and $f\left(x\right)$ has a remainder of $8$ when divided by $x+1$, the value of $k$ is

Answer 1

$f\left(x\right)=(x-3)(x+3)(x+k)$ $(\because 3,-3,-k$ are roots)

$f\left(x\right)=(x^2-9)(x+k)$

$f\left(x\right)=x^3+k{x}^2-9x-9k$

$f\left(x\right)$ has remainder 8 when divided

by $x+1$

$\Rightarrow f(-1)=8$

$f(-1)=-1+k+9-9k$

$8=8-8k$

$0=-8k$

$0=k$

$\therefore k=0$