# Problem Study (Polynomial Division)

1.   If $x^{2}+ax+1$ is a factor of $2x^{3}-16x+b$, Find the values of $\inline&space;a$ and $\inline&space;b$
Solution
Let  $\inline&space;f(x)=2x^3-16x+b$.
Since $x^{2}+ax+1$ is a factor of $\inline&space;f(x)$, the remainder when $\inline&space;f(x)$ is divided by $x^{2}+ax+1$ is 0.
By polynomial long division we can find the remainder.

Hence we have $(2a^2&space;-&space;18)x&space;+&space;2a&space;+&space;b&space;=0$
$\inline&space;\therefore&space;2a^2&space;-&space;18=0$
$\inline&space;\therefore&space;a^2=9$ and $\inline&space;a=\pm&space;3$.
Similarly, we can say $\inline&space;2a+b=0$,$\inline&space;b=-2a$
When $\inline&space;a=-3$, $\inline&space;b=6$ and
When $\inline&space;a=3$, $\inline&space;b=-6$.
-------------------------------------------------------------------------
2.   If $x^3+mx+n$ is divisible by $\inline&space;(x-k)^2$, prove that $\left&space;(\frac{&space;m}{3}&space;\right&space;)^3+\left&space;(&space;\frac{n}{2}&space;\right&space;)^2=0$.
Solution
Since $x^3+mx+n$ is divisible by $\inline&space;(x-k)^2$, The remainder when $x^3+mx+n$ is divided by $\inline&space;(x-k)^2$ = $\inline&space;x^2-2kx+k^2$ is zero.
By polynomial long division,

$\therefore&space;m=-3k^2$ and $n=2k^3$.
$\therefore&space;\left&space;(\frac{&space;m}{3}&space;\right&space;)^3+\left&space;(&space;\frac{n}{2}&space;\right&space;)^2=\left&space;(\frac{&space;-3k^2}{3}&space;\right&space;)^3+\left&space;(&space;\frac{2k^3}{2}&space;\right&space;)^2=-k^6+k^6=0$.