Problem Study (Polynomial Division)

16 Jun, 2016 Posting Komentar

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

Hence we have $(2a^2 - 18)x + 2a + b =0$ .
$\therefore 2a^2 - 18=0$
$\therefore a^2=9$ and $a=\pm 3$ .
Similarly, we can say $2a+b=0$ , $b=-2a$
When $a=-3$ , $b=6$ and
When $a=3$ , $b=-6$ .

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2. If $x^3+mx+n$ is divisible by $(x-k)^2$ , prove that $\left (\frac{ m}{3} \right )^3+\left ( \frac{n}{2} \right )^2=0$ .
Solution
Since $x^3+mx+n$ is divisible by $(x-k)^2$ , The remainder when $x^3+mx+n$ is divided by $(x-k)^2$ = $x^2-2kx+k^2$ is zero.
By polynomial long division,

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

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Problem Study (Polynomial Division)

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