Question

Using factor theorem, show that g(x) is a factor of p(x), when
$p\left(x\right)=2\sqrt{2}{x}^2+5x+\sqrt{2},g\left(x\right)=x+\sqrt{2}$

Answer 1

Let:
$p\left(x\right)=2\sqrt{2}{x}^2+5x+\sqrt{2}$
Here,
$x+\sqrt{2}=0\Rightarrow x=-\sqrt{2}$
By the factor theorem, $\left(x+\sqrt{2}\right)$ will be a factor of the given polynomial if $p\left(-\sqrt{2}\right)$ = 0.
Thus, we have:
$$\begin{gathered} p\left(-\sqrt{2}\right)=\left[2\sqrt{2}\times {\left(-\sqrt{2}\right)}^2+5\times \left(-\sqrt{2}\right)+\sqrt{2}\right] \\ =\left(4\sqrt{2}-5\sqrt{2}+\sqrt{2}\right) \\ =0 \end{gathered}$$
Hence, $\left(x+\sqrt{2}\right)$ is a factor of the given polynomial.