Question

The value of p for which the polynomial $x^3+4x^2-px+8$ is exactly divisible by $(x-2)$ is

• A. 0
• B. 3
• C. 5
• D. 16

Answer 1

The correct option is D 16

Let the function be represented by $f\left(x\right)$.

As $x-2$, exactly divides $f\left(x\right)$, $2$ is the zero of f(x)

Hence, $f\left(2\right)=0$.

$f\left(x\right)=x^3+4x^2-px+8$

$\Rightarrow f\left(2\right)=2^3+4\times 2^2-p\times 2+8=0$

$\Rightarrow 2p=32$

$\Rightarrow p=16$