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Factor the polyomial completely using any method r^5+r^3-r^2-1 Medal given for answer
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\[r^5+r^3-r^2+1\]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\[=(r^2-1)(r^3-1)\]
thanks medal given :)
this problem is not yet completely factored; look at you original problem is the last term supposed to be POSITVE?
no its right
r^3(r^2 + 1) - (r^2 + 1) (r^3 - 1)(r^2 + 1)
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furthermore, (-1+r) (1+r+r^2)(r^2 + 1)
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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