Factor completely

80b^2+200bf+125f^2

took out the common factor of 5

5(16b^2+40bf+25f^2)

am I finish at this point?or what do I do next.

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- Aug 27th 2009, 08:39 AMBrama BullHelp with what to do next
Factor completely

80b^2+200bf+125f^2

took out the common factor of 5

5(16b^2+40bf+25f^2)

am I finish at this point?or what do I do next. - Aug 27th 2009, 08:51 AMVonNemo19
You are NOT done because factors.

Let's say that after taking out the 5 you ended up with

.

you could factor again getting*Then*

.

I say this just so that in the future you will be ready, because the day is coming where you will apply this knowlege quite frequently.

So with that in mind

- Aug 27th 2009, 08:51 AMWilmer
Hint: 16b^2 + 40bf + 25f^2 = (?b + ?f)^2

- Aug 27th 2009, 09:33 AMBrama Bull
(?b + ?f)^2

I am confused now, what happens to the 5 from this

5(16b^2+40bf+25f^2)

I have know clue what is going on with this problem? - Aug 27th 2009, 09:38 AMmathaddict
- Aug 27th 2009, 09:45 AMWilmer
- Aug 27th 2009, 09:46 AMVonNemo19
- Aug 27th 2009, 09:48 AMBrama Bull
(Headbang) yessssssssss, I think I understand slightly what you are trying to tell me with the a^2 +2ab + b^2=(a+b)^2

because that factors out to be what have, I got confused because I forgot about the formula

5(4b+5f)^2

now, is this problem finally complete? if I still haven't got it thanks for your help anyway.(Clapping) - Aug 27th 2009, 09:50 AMBrama Bull
- Aug 27th 2009, 09:52 AMVonNemo19
- Aug 27th 2009, 10:08 AMVonNemo19
Yes. The problem is done.

Watch this

Can you see that if I wanted to factor all I would have to realize is that the first term is squred, the last term is squared, and the middle term is twice the product of the squre roots of the first and last terms.

So, looking at your problem

So can you see how this factors now?