Question

If $x^3+m{x}^2+nx+6$ has $(x-2)$ as factor and leaves a remainder $3$ when divided by $(x-3)$ find the values of $m,n$

• A. $m=2,n=2$
• B. $m=2,n=-2$
• C. $m=-2,n=1$
• D. $m=-3,n=-1$

Answer 1

Answer: (D)

$(m=-3,n=-1)$

Given polynomial be $f\left(x\right)=x^3+m{x}^2+nx+6$ and its factor is $(x-2)$
So, $x=2$ is root of the given polynomial.
Thus, $f\left(2\right)=0$
$\Rightarrow 2^3+m\left(2^2\right)+n\left(2\right)+6=0$
$8+4m+2n+6=0$
$4m+2n=-14$........(i)

And also we have, when $f\left(x\right)$ divided by $x-3$ gives remainder $3$.

$\Rightarrow f\left(3\right)=3$

$\Rightarrow 33+9m+3n=3$

$\Rightarrow 3m+n=-10$.........(ii)
from equation(i)$-2\times$eqation(ii), we get
$4m+2n-6m-2n=-14+20$
$-2m=6$
$\therefore m=-3$
Now, substitute $m=-3$ in equation (i), we get
$4(-3)+2n=-14$
$-12+2n=-14$
$2n=-14+12$

$2n=-2$
$\therefore n=-1$
Hence, $m=-3$ and $n=-1$.