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, 07:39 AM #1

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- Aug 27th 2009, 07:51 AM #2
You are NOT done because $\displaystyle 16b^2+40bf+25f^2$ factors.

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

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

you could factor again getting*Then*

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

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

$\displaystyle 16^2+40bf+25f^2=(4b+5f)(4b+5f)$

- Aug 27th 2009, 07:51 AM #3

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- Aug 27th 2009, 08:33 AM #4

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- Aug 27th 2009, 08:38 AM #5

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Well for this part ..

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

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

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

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

Am i confusing you further ?

- Aug 27th 2009, 08:45 AM #6

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- Aug 27th 2009, 08:46 AM #7

- Aug 27th 2009, 08:48 AM #8

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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.

- Aug 27th 2009, 08:50 AM #9

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- Aug 27th 2009, 08:52 AM #10

- Aug 27th 2009, 09:08 AM #11
Yes. The problem is done.

Watch this

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

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

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

So can you see how this factors now?