Question

For what value of $m+14$ is $2x^3+m{x}^2+11x+m+3$ exactly divisible by $(2x-1)$?

Answer 1

Let $f\left(x\right)=2x^3+m{x}^2+11x+m+3$ and

$g\left(x\right)=2x-1$

$=>2x-1=0$

$=>x=\frac{1}{2}$

If $f\left(x\right)$ is exactly divisible by $g\left(x\right)$ then $f\left(\frac{1}{2}\right)=0$ (By Remainder Theorem)

Thus, $f\left(\frac{1}{2}\right)=0$

$=>2x^3+m{x}^2+11x+m+3=0$

$=>2(\frac{1}{2}{)}^3+m(\frac{1}{2}{)}^2+11\left(\frac{1}{2}\right)+m+3=0$

$=>\frac{1}{4}+\frac{m}{4}+\frac{11}{2}+m+3=0$

$=>\frac{1+m+22+4m+12}{4}=0$

$=>\frac{5m+35}{4}=0$

$=>5m+35=0$

$=>5m=-35$

$=>m=-\frac{35}{5}$

$=>m=-7$