Question

Factorize:
$\sqrt{2}{x}^2+3x+\sqrt{2}$

Answer 1

We have:
$\sqrt{2}{x}^2+3x+\sqrt{2}$
We have to split 3 into two numbers such that their sum is 3 and their product is 2, i.e., $\sqrt{2}\times \sqrt{2}$.
Clearly, $2+1=3\mathrm{and}2\times 1=2$.

$$\begin{gathered} \therefore \sqrt{2}{x}^2+3x+\sqrt{2}=\sqrt{2}{x}^2+2x+x+\sqrt{2} \\ =\sqrt{2}x\left(x+\sqrt{2}\right)+1\left(x+\sqrt{2}\right) \\ =\left(x+\sqrt{2}\right)\left(\sqrt{2}x+1\right) \end{gathered}$$