Find the value of a for which the polynomial

Solution:

Let:

$f(x)=x^{4}-x^{3}-11 x^{2}-x+a$

Now, 

$x+3=0 \Rightarrow x=-3$

By the factor theorem, $f(x)$ is exactly divisible by $(x+3)$ if $f(-3)=0$.

Thus, we have:

$f(-3)=\left[(-3)^{4}-(-3)^{3}-11 \times(-3)^{2}-(-3)+a\right]$

$=(81+27-99+3+a)$

$=12+a$

Also

$f(-3)=0$

$\Rightarrow 12+a=0$

$\Rightarrow a=-12$

Hence, $f(x)$ is exactly divisible by $(x+3)$ when $a$ is $-12$.