Question

For what value of $m$ is $x^3-2m{x}^2+16$ divisible by $x+2$?

Answer 1

Step 1: Calculate $g\left(x\right)$

Let $p\left(x\right)=x^3-2m{x}^2+16$,

$g\left(x\right)=x+2$

$g\left(x\right)=0$

$\therefore x+2=0$

$\Rightarrow$ $x=–2$

Therefore, zero of $g\left(x\right)=–2$.

We know that, according to factor theorem, if $f\left(x\right)$ is a polynomial of degree $n\ge 1$ and ‘$a$’ is any real number, then, $(x-a)$ is a factor of $f\left(x\right)$, if $f\left(a\right)=0$.

if $p\left(x\right)$ is divisible by $g\left(x\right)$, then the remainder $p(-2)$ should be zero.

Step 2: Calculate the value of $m$

On substituting the value of $x$ in $p\left(x\right)$, we obtain

$p(–2)=0$

$\Rightarrow$${(–2)}^3–2m{(–2)}^2+16=0$

$\Rightarrow$ $0–8–8m+16=0$

$\Rightarrow$ $8m=8$

$\Rightarrow$ $m=1$

Hence, for $m=1$, $x^3-2m{x}^2+16$ is divisible by $x+2$.