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Mathematics 7 Online
OpenStudy (anonymous):

What does 13m2n3p + 21mp2 equal when completely factored? (A)13mp(mn3 + 2p) (B)m2p2(13n3 + 21) (C)mp(13mn3 + 21p) (D)mnp(13mn2 + 21p)

OpenStudy (anonymous):

i know this one!

OpenStudy (abb0t):

It looks like you can factor out an m and p.

OpenStudy (anonymous):

let me do it!

OpenStudy (anonymous):

dude!

OpenStudy (abb0t):

Also. combine all numbers since they are just constants. So multiply 13 x 2 x 3 :)

OpenStudy (anonymous):

um nvm i dont no this

OpenStudy (abb0t):

Put constants and variables together but keep them separate so you can see

OpenStudy (anonymous):

idk i think so

OpenStudy (anonymous):

let meh try

OpenStudy (abb0t):

wait, nevermind, are those 3 and 2's mean they are raised to that power?

OpenStudy (anonymous):

|dw:1365966144717:dw|

OpenStudy (anonymous):

sry bad at this i am young ok

OpenStudy (anonymous):

abb0t yes the 2 and 3 are raised to the power

OpenStudy (abb0t):

\(13m^2n^3p +21m p^2\) is that correct?

OpenStudy (abb0t):

Ok, simply factor the lowest common factor between them :)

OpenStudy (anonymous):

Yes that is correct

OpenStudy (abb0t):

Your two common variables are "m" and "p" That's because you have both of them there. There's no "n" variable on the right side. Now, notice that you have ONE "p" on the left and ONE "m" on the left. Those are your common factors.

OpenStudy (abb0t):

There's also no two common factors between 13 and 21, so you can just leave them in there. In essence, you're taking out ONE "m" and ONE "p" only.

OpenStudy (abb0t):

Now, can you make what your answer is?

OpenStudy (anonymous):

would it be C?

OpenStudy (nurali):

correct

OpenStudy (abb0t):

Yep.

OpenStudy (anonymous):

thank you

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