Solve this

(b) $m=7, n=-18$

Let:

$p(x)=x^{3}+10 x^{2}+m x+n$

Now,

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

$(x+2)$ is a factor of $p(x)$.

So, we have $p(-2)=0$

$\Rightarrow(-2)^{3}+10 \times(-2)^{2}+m \times(-2)+n=0$

$\Rightarrow-8+40-2 m+n=0$

$\Rightarrow 32-2 m+n=0$

$\Rightarrow 2 m-n=32$ ........(i)

Now,

$x-1=0 \Rightarrow x=1$

Also, 

$(x-1)$ is a factor of $p(x)$.

We have:
p(1) = 0

$\Rightarrow 1^{3}+10 \times 1^{2}+m \times 1+n=0$

$\Rightarrow 1+10+m+n=0$

$\Rightarrow 11+m+n=0$

$\Rightarrow m+n=-11 \quad \ldots$ (ii)

From (i) and (ii), we get:

$3 m=21 \Rightarrow m=7$

By substituting the value of $m$ in (i), we get $n=-18$.

$\therefore m=7$ and $n=-18$