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Mathematics
OpenStudy (anonymous):

factor x^2 - 3x

OpenStudy (anonymous):

how would you do that ?

OpenStudy (anonymous):

take out at term that both sets have in common

OpenStudy (anonymous):

a term*

OpenStudy (anonymous):

so x

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

exactly! so try factoring it out ^.^

OpenStudy (anonymous):

x^2-3?

OpenStudy (anonymous):

you have to take the x out of both parts

OpenStudy (anonymous):

this really looks like (x*x-3x)

OpenStudy (anonymous):

so what will it look like ?

OpenStudy (anonymous):

so in what you just put, you factored the x out of the second part, but not the first. factoring it out doesn't mean that it disappears, it just means that you take it out of the equation.. think of it as the opposite of distribution.. so if you're distributing x(5x+3) you get \[5x^2+3x\] when you factor that x out, you go right back to x(5x+3)

OpenStudy (anonymous):

so for my problem i would x * x-3x?

OpenStudy (anonymous):

yes :)

OpenStudy (anonymous):

Thanks!

OpenStudy (anonymous):

almost! you're close in both of the answers you put you have half of the answer correct.. try distributing what you just wrote out again (and be sure to put parenthesis to indicate that you're multiplying the entire equation, and not just the first x

OpenStudy (anonymous):

what you just put is x(x-3x), if you distribute it, its actually \[x*2-3x^2\]

OpenStudy (anonymous):

sorry i misread your answer @itzmashy is correct

OpenStudy (anonymous):

oops, sorry formatting messed upt, it's actually: \[x^2-3x^2\]

OpenStudy (anonymous):

try factoring out the x, one more time

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