Question

Find the value of m and n when the polynomial $\text{f(x)}={\text{x}}^3-2{\text{x}}^2+\text{mn}+\text{mn}$ has a factor $(\text{x}+2)$ and leaves a remainder 9 when divided by $(\text{x}+1)$.​

Answer 1

Step 1: Form the equations
Given: $\text{f(x)}={\text{x}}^3-2{\text{x}}^2+\text{mn}+\text{mn}$

A polynomial $\text{f(x)}$ has a factor $\text{(x-a)}$ if and only if $\text{f(a)}=0$.

f(x) has a factor (x + 2).
$\therefore \text{f(-2)}=0$
$(-2{)}^3-2(-2{)}^2+\text{m}(-2)+\text{n}=0$
$-8-8-2\text{m}+\text{n}=0$
$\text{n}=2\text{m}+16$-(i)

Also, $\text{f(x)}$ leaves remainder $9$ when divided by $\text{(x + 1)}$
$\therefore$By remainder theorem, when a polynomial $\text{f(x)}$ is divided by $\text{f(x-a)}$, then the remainder is $\text{f(a)}$.

$\therefore \text{f(-1)}$=9
$(-1{)}^3-2(-1{)}^2+\text{m}(-1)+\text{n}=9$
$-1-2-\text{m}+\text{n}=9$
$\text{n}-\text{m}=12$-(ii)

Step 2: Find the value of m and n
Put the value of n in equation (ii)
$2\text{m}+16-\text{m}=12$
$\text{m}=-4$

Put the value of m in equation(ii)
$\text{n}=2(-4)+16$
$\text{n}=8$

Hence, $\text{m}=-4$ and $\text{n}=8$ .