Question

Use the factor theorem to determine whether $g\left(x\right)$ is a factor of $f\left(x\right)$

$f\left(x\right)=2\sqrt{2}{x}^2+5x+\sqrt{2};g\left(x\right)=x+\sqrt{2}$

Answer 1

Checking whether $g\left(x\right)$ is a factor of $f\left(x\right)$ or not.

According to the factor theorem, if $g\left(x\right)=(x-a)$ is a factor of $f\left(x\right)$ then $f\left(a\right)=0$.

Here, $f\left(x\right)=2\sqrt{2}{x}^2+5x+\sqrt{2};g\left(x\right)=x+\sqrt{2}$

Now, substitute $-\sqrt{2}$ for $x$ in the function $f\left(x\right)$:

$$\begin{gathered} f(-\sqrt{2})=2\sqrt{2}{(-\sqrt{2})}^2+5(-\sqrt{2})+\sqrt{2} \\ \Rightarrow f(-\sqrt{2})=4\sqrt{2}-5\sqrt{2}+\sqrt{2} \\ \therefore f(-\sqrt{2})=0 \end{gathered}$$

As $f(-\sqrt{2})=0$, the given polynomial $g\left(x\right)$ is a factor of $f\left(x\right)$.

Hence, the polynomial $g\left(x\right)=x+\sqrt{2}$ is a factor of $f\left(x\right)=2\sqrt{2}{x}^2+5x+\sqrt{2}$