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Mathematics 8 Online
OpenStudy (anonymous):

Factor the polyomial completely using any method r^5+r^3-r^2-1 Medal given for answer

OpenStudy (anonymous):

\[r^5+r^3-r^2+1\]factor out the GCF from the first two terms, and the -1 from the second two\[=r^3(r^2-1)-(r^2-1)\]factor out the binomial\[=(r^2-1)(r^3-1)\]

OpenStudy (anonymous):

thanks medal given :)

OpenStudy (anonymous):

this problem is not yet completely factored; look at you original problem is the last term supposed to be POSITVE?

OpenStudy (anonymous):

no its right

OpenStudy (ybarrap):

r^3(r^2 + 1) - (r^2 + 1) (r^3 - 1)(r^2 + 1)

OpenStudy (ybarrap):

furthermore, (-1+r) (1+r+r^2)(r^2 + 1)

OpenStudy (anonymous):

then my last line is incorrect it should be \[=r^3(r^2+1)-(r^2+1)\]\[=(r^2+1)(r^3-1)\]the second binomial is a difference of cubes\[=(r^2+1)(r-1)(r^2+r+1)\]and this is the factored form.

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