Question

If $x^3+ax-28$ is exactly divisible by $x-4$, then the value of $a$ is

• A. $23$
• B. $-23$
• C. $9$
• D. $-9$

Answer 1

The correct option is D $-9$

$x^3+\left(ax\right)-28$ is exactly divisible by $(x-4)$ which means $4$ is a zero of $x^3+ax-28$

$\therefore {\left(4\right)}^3+a\left(4\right)-28=0$

$\Rightarrow 64+4a-28=0$

$\Rightarrow 4a=-36$

$\Rightarrow a=-9.$

Hence, the answer is $-9.$