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

2 Answers
Feb 21, 2017

Answer:

#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)#

Feb 21, 2017

Answer:

#(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.