Using factor theorem, show that g(x) is a factor of p(x), when

Let:

$p(x)=2 x^{4}+x^{3}-8 x^{2}-x+6$

Here, 

$2 x-3=0 \Rightarrow x=\frac{3}{2}$

By the factor theorem, $(2 x-3)$ is a factor of the given polynomial if $p\left(\frac{3}{2}\right)=0$.

Thus, we have:

$p\left(\frac{3}{2}\right)=\left[2 \times\left(\frac{3}{2}\right)^{4}+\left(\frac{3}{2}\right)^{3}-8 \times\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)+6\right]$

$=\left(\frac{81}{8}+\frac{27}{8}-18-\frac{3}{2}+6\right)$

$=0$

Hence, (2x -">−- 3) is a factor of the given polynomial.