Mathematics
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OpenStudy (anonymous):
Factor This:
3x^3y-6x^2y^2+3xy^3
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OpenStudy (xapproachesinfinity):
Factor 3xy first as a start
OpenStudy (anonymous):
3 x (-2 x + x^2 y - y^3) @Purpleloverr
OpenStudy (xapproachesinfinity):
No straight answer, read open study code of conduct
OpenStudy (xapproachesinfinity):
In fact it's not even correct
OpenStudy (anonymous):
Im confused
D:
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OpenStudy (xapproachesinfinity):
What is the common factor between all of the terms
OpenStudy (anonymous):
so first factor 3xy?
OpenStudy (anonymous):
3xy
OpenStudy (xapproachesinfinity):
Yah because that's the common factor
OpenStudy (anonymous):
yes i got that
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OpenStudy (xapproachesinfinity):
OK factor that what will happen
OpenStudy (anonymous):
I got 3xy(x^2-2xy+y^2)
OpenStudy (xapproachesinfinity):
Good
OpenStudy (anonymous):
i dont see any other common factor
OpenStudy (xapproachesinfinity):
Now yhave a quadratic of the form a^2-2ab+b^2
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OpenStudy (anonymous):
so treat it as a plug in?
OpenStudy (anonymous):
I am confused
OpenStudy (xapproachesinfinity):
No that's just a reminder for that it is of the same type
OpenStudy (anonymous):
so do you use the quadratic formula?
OpenStudy (xapproachesinfinity):
How do you factor a^2-2ab+b^2 you should know this
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OpenStudy (xapproachesinfinity):
No binomial factorization.
OpenStudy (xapproachesinfinity):
We know that a^2-2ab+b^2=(a-b)^2
OpenStudy (xapproachesinfinity):
You will do the same to that
Instead of a and b you have x and y
OpenStudy (anonymous):
Ok i see, sorry i was sick for the past 2 weeks so i missed a lot...
OpenStudy (xapproachesinfinity):
it's ok ^_-
i was on the phone so i couldn't explain further :)