# Help with what to do next

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• Aug 27th 2009, 07:39 AM
Brama Bull
Help 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, 07:51 AM
VonNemo19
Quote:

Originally Posted by Brama Bull
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.

You are NOT done because $16b^2+40bf+25f^2$ factors.

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

$5(b^2+2bf+f^2)$.

Then you could factor again getting

$5(b+f)(b+f)$.

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

$16^2+40bf+25f^2=(4b+5f)(4b+5f)$
• Aug 27th 2009, 07:51 AM
Wilmer
Hint: 16b^2 + 40bf + 25f^2 = (?b + ?f)^2
• Aug 27th 2009, 08:33 AM
Brama 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, 08:38 AM
mathaddict
Quote:

Originally Posted by Brama 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?

Well for this part ..

$16b^2+40bf+25f^2$

it has the form of $a^2+2ab+b^2=(a+b)^2$

in this case $a^2=16b^2$ and $b^2=25f^2$ , a=4b , b=5f

so $(4b)^2+2(4b)(5f)+(5f)^2=(4b+5f)^2$

Am i confusing you further ? (Doh)
• Aug 27th 2009, 08:45 AM
Wilmer
Quote:

Originally Posted by Brama 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?

Geezzzz BB, just write it in: 5(?b + ?f)^2
• Aug 27th 2009, 08:46 AM
VonNemo19
Quote:

Originally Posted by Brama 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?

Did my last post not help?
• Aug 27th 2009, 08:48 AM
Brama 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, 08:50 AM
Brama Bull
Quote:

Originally Posted by Wilmer
Geezzzz BB, just write it in: 5(?b + ?f)^2

I like to understand what is going on that is why I ask questions.
• Aug 27th 2009, 08:52 AM
VonNemo19
Quote:

Originally Posted by Wilmer
Geezzzz BB, just write it in: 5(?b + ?f)^2

Don't make the OP feel like he should have seen what you said. I think he's having trouble visualizing this. Do you have another way of explaining the factorization of trinomials?
• Aug 27th 2009, 09:08 AM
VonNemo19
Quote:

Originally Posted by Brama 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)

Yes. The problem is done.

Watch this

$(x+y)(x+y)=x^2+2xy+y^2$

Can you see that if I wanted to factor $x^2+2xy+y^2$ 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

$\overbrace{16b^2}^{\text{first term is squared}}+\overbrace{40bf}^{\text{twice the product}}+\overbrace{25f^2}^{\text{last term is squared}}$

So can you see how this factors now?