I don't know how you're meant to factor those, but wolfie says
\[\Large (x-1)(x^2+2)^2\]
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OpenStudy (anonymous):
eh? how wolfie get that?
OpenStudy (fifciol):
\[x^4(x-1)+4x^2(x-1)+4(x-1)\]
OpenStudy (anonymous):
how??
OpenStudy (fifciol):
now factor x-1
terenzreignz (terenzreignz):
It's called factoring by grouping -.-'
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OpenStudy (anonymous):
@Fifciol .. ahh.. got it. :) use synthetic division??
OpenStudy (unklerhaukus):
lets look at the denominator first
\[x^5-x^4+4x^3-4x^2+4x-4\]
lefts factor the four out
\[x^5-x^4+4(x^3-x^2+x-1)\]
now lets look at this term
\[x^3-x^2+x-1\]
OpenStudy (unklerhaukus):
*lets
OpenStudy (unklerhaukus):
\[x^3-x^2+x-1\]
we can try polynomial long division long division|dw:1375784983206:dw|