Find the values of a and b so that the polynomial

Solution:

Let:

$f(x)=x^{4}+a x^{3}-7 x^{2}-8 x+b$

Now,

$x+2=0 \Rightarrow x=-2$

By the factor theorem, we can say:

$f(x)$ will be exactly divisible by $(x+2)$ if $f(-2)=0$.

Thus, we have:

$f(-2)=\left[(-2)^{4}+a \times(-2)^{3}-7 \times(-2)^{2}-8 \times(-2)+b\right]$

$=(16-8 a-28+16+b)$

$=(4-8 a+b)$

$\therefore f(-2)=0 \Rightarrow 8 a-b=4 \quad \ldots(1)$

Also,

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

By the factor theorem, we can say:

$f(x)$ will be exactly divisible by $(x+3)$ if $f(-3)=0$.

Thus, we have:

$f(-3)=\left[(-3)^{4}+a \times(-3)^{3}-7 \times(-3)^{2}-8 \times(-3)+b\right]$

$=(81-27 a-63+24+b)$

$=(42-27 a+b)$

$\therefore f(-3)=0 \Rightarrow 27 a-b=42 \ldots(2)$

Subtracting (1) from (2), we get :

$\Rightarrow 19 a=38$

$\Rightarrow a=2$

Putting the value of a, we get the value of b, i.e., 12.

$\therefore a=2$ and $b=12$