Question

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

Answer 1

`2 d^2 g + 2 f g + 4 d^2 h + 4 f h=2 (d^2 + f) (g + 2 h)`

Explanation:

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

then

`2 (lambda g + f g + 2 lambda h + 2 f h)` appears clearly as a biliear product

`2(lambda +f)(g+2h)`

Answer 2

`(2g+4h)(d^2+f)`

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!

`color(blue)("Try 1")`

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

`(2d^2g+2fg)+(4d^2h+4fh)`

looking for common factors & considering the first brackets gives:
`2g(d^2+f)+....` so if we can get `(d^2+f)` from the second brackets we have our repeats.

`color(green)(color(blue)(2g)(d^2+f)color(blue)(+4h)(d^2+f))`

This is the same as:

`color(blue)((2g+4h))color(green)((d^2+f))`

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