Question

Using factor theorem, show that g(x) is a factor of p(x), when
$p\left(x\right)=7x^2-4\sqrt{2}x-6,g\left(x\right)=x-\sqrt{2}$

Answer 1

Let:
$p\left(x\right)=7x^2-4\sqrt{2}x-6$
Here,
$x-\sqrt{2}=0\Rightarrow x=\sqrt{2}$
By the factor theorem, $\left(x-\sqrt{2}\right)$ is a factor of the given polynomial if $p\left(\sqrt{2}\right)=0$
Thus, we have:
$$\begin{gathered} p\left(\sqrt{2}\right)=\left[7\times {\left(\sqrt{2}\right)}^2-4\sqrt{2}\times \sqrt{2}-6\right] \\ =\left(14-8-6\right) \\ =0 \end{gathered}$$
Hence, $\left(x-\sqrt{2}\right)$ is a factor of the given polynomial.