If $(x+2)$ and $(x-1)$ are factors of $\left(x^{3}+10 x^{2}+m x+n\right)$, then
(a) $m=5, n=-3$
(b) $m=7, n=-18$
(c) $m=17, n=-8$
(d) $m=23, n=-19$
(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 $n=-18$
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