Ask your own question, for FREE!
Algebra 10 Online
OpenStudy (anonymous):

factor 4p^2-10p-50

OpenStudy (anonymous):

First factor out the GCF (greatest common factor). Do you know how to do that?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

that would come out to 2(2p^2-5p-25)

OpenStudy (anonymous):

Good. Then there are lots of different methods for factoring that new trinomial. Do you know what method you're familiar with? Guess and check, box and diamond, bottoms up...? Any of those sound familiar?

OpenStudy (anonymous):

no

OpenStudy (jhannybean):

bottoms up? Haha.

OpenStudy (anonymous):

Bottoms Up is actually a really cool method for factoring trinomials when a>1. It's a funny name, but it actually makes it much easier!

OpenStudy (anonymous):

@candy123456 do you have a lot of these to factor? Or just this one?

OpenStudy (anonymous):

like 4

OpenStudy (anonymous):

Do you want me to show you the bottoms up method? Or do you want to just guess/check these problems?

OpenStudy (anonymous):

whatever is easier

OpenStudy (anonymous):

I teach Algebra 2 -- most of my kids prefer the bottoms up method. Once you get the process, it's definitely easier. But if you understand the method behind factoring and FOILing, then guess/check may be quicker. It just depends on what you're comfortable with.

OpenStudy (anonymous):

bottoms up sounds good to me

OpenStudy (anonymous):

Ok...let me know if you get confused. :)

OpenStudy (anonymous):

First step is multiply a by c. So your problem is a = 2, and c = -25. So slide the 2 over and multiply it by the -25.

OpenStudy (anonymous):

Sorry, the equation disappeared in there...

OpenStudy (anonymous):

\[2p ^{2}-5p-25\] becomes \[p ^{2}-5p-50\] because we did 2 times the 25...

OpenStudy (anonymous):

With me?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Ok now we factor that like a normal trinomial where a=1. Do you know how to do that one?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Ok let me know what you get. 2 steps after that.

OpenStudy (anonymous):

(p-1-)(p+5)

OpenStudy (anonymous):

sorry (p-10)(p+5)

OpenStudy (anonymous):

Good! Now you divide each of those numbers by the original a (the 2, not the 4 -- this gets a little confusing when you factor out a GCF before starting this method...but use the 2). So then we have: \[(p-\frac{ 10 }{ 2 })(p+\frac{ 5 }{ 2 })\] This is a strange process, I know...but stick with me, almost done.

OpenStudy (anonymous):

Reduce any of the fractions, if possible.

OpenStudy (anonymous):

would u reduce the 10/2

OpenStudy (anonymous):

Yep. So then you have: \[(p-5)(p+\frac{ 5 }{ 2 })\]

OpenStudy (anonymous):

Last step. You "bottoms up" the factor with a fraction in it. That means you take the denominator (the bottom) and move it UP in front of the variable. So (p + 5/2) becomes (2p + 5). I know it's weird, but it works!

OpenStudy (anonymous):

With me?

OpenStudy (anonymous):

yes so i would get (p+5)(2p+5) ?

OpenStudy (anonymous):

Yes! But don't forget to stick the 2 in the very front from when you factored out the GCF. So final answer should be 2(p-5)(2p+5).

OpenStudy (anonymous):

oo yeah !

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

No problem!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!