Question:
For what value of $k$ is the polynomial $p(x)=2 x^{3}-k x+3 x+10$ exactly divisible by $(x+2) ?$
(a) $-\frac{1}{3}$
(b) $\frac{1}{3}$
(c) 3
(d) $-3$
Solution:
(d) −3
Let:
$p(x)=2 x^{3}-k x^{2}+3 x+10$
Now,
$x+2=0 \Rightarrow x=-2$
$p(x)$ is completely divisible by $(x+2)$.
$\therefore p(-2)=0$
$\Rightarrow 2 \times(-2)^{3}-k \times(-2)^{2}+3 \times(-2)+10=0$
$\Rightarrow-16-4 k-6+10=0$
$\Rightarrow-12-4 k=0$
$\Rightarrow 4 k=-12$
$\Rightarrow k=\frac{-12}{4}$
$\Rightarrow k=-3$