Question

Question 19
For what valuel of m is $x^3–2m{x}^2+16$ is divisible by x +2?

Answer 1

Let $p\left(x\right)=x^3-2m{x}^2+16$
Since, p(x) is divisible by (x+2), the remainder = 0
$\therefore p(-2)=0$ [ by remainder theorem]
$\Rightarrow (-2{)}^3–2m(-2{)}^2+16=0$
$\Rightarrow -8–8m+16=0$
$\Rightarrow 8=8m$
$\therefore m=1.$
Hence, the value of m is 1.