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Mathematics 6 Online
OpenStudy (anonymous):

Factor completely, given that (x-2) is a factor. y=x^3-12x^2+36x-32 Please show work!!! Thanks!

OpenStudy (anonymous):

ok you know it has to look like \[(x-2)(\text{something})\] right? and you can find the "something" by dividing

OpenStudy (anonymous):

or you can find the "something" by thinking has to look like \[(x-2)(x^2+bx-16)\]

OpenStudy (anonymous):

actually it has to look like \[(x-2)(x^2+bx+16)\]since \[-2\times 16=-32\]

OpenStudy (anonymous):

b can be found easy enough since \[-2x^2+bx^2=-12x^2\] \[-2+b=-12\] \[b=-10\]

OpenStudy (anonymous):

or if you prefer you could start again with \[(x-2)(x^2+bx+16)\]and then think that \[-2bx+16x=36x\] \[-2b+16=36\] \[-2b=20\] \[b=-10\] either way, doesn't matter which you use

OpenStudy (anonymous):

Thank you :) When it comes to math, i have no idea what to do... O_o lol

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