Question

If (x + 2) and (x − 1) are factors of (x³ + 10x² + mx + n), then
(a) m = 5, n = −3
(b) m = 7, n = −18
(c) m = 17, n = −8
(d) m = 23, n = −19

Answer 1

(b) m = 7, n = −18

Let:
$p\left(x\right)=x^3+10x^2+mx+n$
Now,
$x+2=0\Rightarrow x=-2$
(x + 2) is a factor of p(x).
So, we have p($-$2)=0
$$\begin{gathered} \Rightarrow {\left(-2\right)}^3+10\times {\left(-2\right)}^2+m\times \left(-2\right)+n=0 \\ \Rightarrow -8+40-2m+n=0 \\ \Rightarrow 32-2m+n=0 \\ \Rightarrow 2m-n=32.....\left(\mathrm{i}\right) \end{gathered}$$
Now,
$x-1=0\Rightarrow x=1$
Also,
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0
$$\begin{gathered} \Rightarrow 1^3+10\times 1^2+m\times 1+n=0 \\ \Rightarrow 1+10+m+n=0 \\ \Rightarrow 11+m+n=0 \\ \Rightarrow m+n=-11.....\left(\mathrm{ii}\right) \\ \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}: \\ 3m=21\Rightarrow m=7 \end{gathered}$$
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18