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

Let:

$p(x)=7 x^{2}-4 \sqrt{2} x-6$

Here, 

$x-\sqrt{2}=0 \Rightarrow x=\sqrt{2}$

By the factor theorem, $(x-\sqrt{2})$ is a factor of the given polynomial if $p(\sqrt{2})=0$

Thus, we have:

$p(\sqrt{2})=\left[7 \times(\sqrt{2})^{2}-4 \sqrt{2} \times \sqrt{2}-6\right]$

$=(14-8-6)$

$=0$

Hence, $(x-\sqrt{2})$ is a factor of the given polynomial.