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 $p(x)=x^{3}-20 x+5 k$.
$\Rightarrow(-5)^{3}-20 \times(-5)+5 k=0$
$\Rightarrow-125+100+5 k=0$
$\Rightarrow 5 k=25$
$\Rightarrow k=5$
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