Question

Question 21
Find the value of m, so the 2x- 1 be a factor of $8x^4+4x^3–16x^2+10x+m$.

Answer 1

Let $p\left(x\right)=8x^4+4x^3-16x^2+10x+m$.
Since 2x - 1 is a factor of p(x), $P\left(\frac{1}{2}\right)=0$[Using Remainder Theorem]
$\therefore 8{\left(\frac{1}{2}\right)}^4+4{\left(\frac{1}{2}\right)}^3-16{\left(\frac{1}{2}\right)}^2+10\left(\frac{1}{2}\right)+m=0$
$\Rightarrow 8\times \frac{1}{16}+4\times \frac{1}{8}-16\times \frac{1}{4}+10\left(\frac{1}{2}\right)+m=0$
$\Rightarrow \frac{1}{2}+\frac{1}{2}-4+5+m=0$
$\therefore 1+1+m=0$
$m=-2$