Find the value of m for which (2x – 1) is a factor of
Question:

Find the value of $m$ for which $(2 x-1)$ is a factor of $\left(8 x^{4}+4 x^{3}-16 x^{2}+10 x+m\right)$.

Solution:

Let $f(x)=8 x^{4}+4 x^{3}-16 x^{2}+10 x+m$

It is given that $(2 x-1)=2\left(x-\frac{1}{2}\right)$ is a factor of $f(x)$.

Using factor theorem,  we have

$f\left(\frac{1}{2}\right)=0$

$\Rightarrow 8 \times\left(\frac{1}{2}\right)^{4}+4 \times\left(\frac{1}{2}\right)^{3}-16 \times\left(\frac{1}{2}\right)^{2}+10 \times \frac{1}{2}+m=0$

$\Rightarrow \frac{1}{2}+\frac{1}{2}-4+5+m=0$

$\Rightarrow 2+m=0$

$\Rightarrow m=-2$

Thus, the value of $m$ is $-2$.