If (x + 5) is a factor of p(x)

Question:

If $(x+5)$ is a factor of $p(x)=x^{3}-20 x+5 k$, then $k=?$

(a) $-5$

(b) 5

(c) 3

(d) $-3$

Solution:

(b) 5

$(x+5)$ is a factor of $p(x)=x^{3}-20 x+5 k$.

$\therefore p(-5)=0$

$\Rightarrow(-5)^{3}-20 \times(-5)+5 k=0$

$\Rightarrow-125+100+5 k=0$

$\Rightarrow 5 k=25$

$\Rightarrow k=5$