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

Let:

$p(x)=2 \sqrt{2} x^{2}+5 x+\sqrt{2}$

Here,

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

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

Thus, we have:

$p(-\sqrt{2})=\left[2 \sqrt{2} \times(-\sqrt{2})^{2}+5 \times(-\sqrt{2})+\sqrt{2}\right]$

$=(4 \sqrt{2}-5 \sqrt{2}+\sqrt{2})$

$=0$

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