# Thread: Help with what to do next

1. ## 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.

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

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

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

Then you could factor again getting

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

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

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

3. Hint: 16b^2 + 40bf + 25f^2 = (?b + ?f)^2

4. (?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?

5. 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 ..

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

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 ?

6. 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

7. 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?

8. 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.

9. 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.

10. 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?

11. Originally Posted by Brama Bull
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.
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.

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