# How do you factor 2d^2g+2fg+4d^2h+4fh?

Feb 21, 2017

$2 {d}^{2} g + 2 f g + 4 {d}^{2} h + 4 f h = 2 \left({d}^{2} + f\right) \left(g + 2 h\right)$

#### Explanation:

$d$ appears always to the second power. so we can call it as $\lambda$

then

$2 \left(\lambda g + f g + 2 \lambda h + 2 f h\right)$ appears clearly as a biliear product

$2 \left(\lambda + f\right) \left(g + 2 h\right)$

Feb 21, 2017

$\left(2 g + 4 h\right) \left({d}^{2} + f\right)$

#### Explanation:

Sometimes the associations can be spotted more easily than at other times.

You are looking for repeats. Notice that there are two ${d}^{2}$ implying we could have the format of: ?(d^2+?)+?(d^2+?)

So lets experiment!

$\textcolor{b l u e}{\text{Try 1}}$

Grouping by coefficients (numbers) so that they can be factored out.

$\left(2 {d}^{2} g + 2 f g\right) + \left(4 {d}^{2} h + 4 f h\right)$

looking for common factors & considering the first brackets gives:
$2 g \left({d}^{2} + f\right) + \ldots .$ so if we can get $\left({d}^{2} + f\right)$ from the second brackets we have our repeats.

$\textcolor{g r e e n}{\textcolor{b l u e}{2 g} \left({d}^{2} + f\right) \textcolor{b l u e}{+ 4 h} \left({d}^{2} + f\right)}$

This is the same as:

$\textcolor{b l u e}{\left(2 g + 4 h\right)} \textcolor{g r e e n}{\left({d}^{2} + f\right)}$

There is no need to experiment further as we have found the answer.