Question

For what value of k is the polynomial $p\left(x\right)=2x^3-k{x}^2+3x+10$ exactly divisible by (x+2)?
(a) $-\frac{1}{3}$
(b) $\frac{1}{3}$
(c) $3$
(d) $-3$

Answer 1

Let $p\left(x\right)=2x^3-k{x}^2+3x+10$
Now,
x+2=0 ⇒x=-2
p(x) is completely divisible by x+2.

∴ p(-2)=0

⇒$2\times (-2{)}^3-k\times (-2{)}^2+3\times -2+10=0$

⇒-16-4k-6+10=0

⇒-12-4k=0

⇒4k=-12

⇒k=$\frac{-12}{4}$

⇒k=-3

So, the correct option is (d).