Question

Using remainder theorem, find the value of m if the polynomial ​
$\text{f(x)}={\text{x}}^3+{\text{5x}}^2-\text{mx}+6$ leaves a remainder $2\text{m}$ when divided by ($\text{x-1}$).

Answer 1

Given: $\text{f(x)}={\text{x}}^3+{\text{5x}}^2-\text{mx}+6$, divisor = $\text{(x-1)}$, and
remainder = $\text{2m}$

From remainder theorem, we know that
when a polynomial $\text{f(x)}$ is divided by $\text{(x-a)}$, the
remainder is $\text{f(a)}$.

$\therefore \text{f(1)}=2m$
$\left(1\right)³+5\left(1\right)²-\text{m}\left(1\right)+6=2\text{m}$
$1+5-\text{m}+6=2\text{m}$
$3\text{m}=12$
$\text{m}=4$

Hence, the value of m is $4$.