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Mathematics 17 Online
OpenStudy (anonymous):

Find all zeros of the polynomial. P(x) = x^5 - 29x^3 - 58x^2 + x - 29.

OpenStudy (anonymous):

this is written correctly? there are no rational roots to help simplify it

OpenStudy (anonymous):

Right, it can be divided with synthetic division by 29

OpenStudy (paxpolaris):

remainder won't be 0

OpenStudy (anonymous):

then factored by (x^2+1)^2, bit it needs to be reduced further

OpenStudy (anonymous):

as pax said, 29 isn't a zero

OpenStudy (anonymous):

unless you made a typo

OpenStudy (anonymous):

i reconstructed the original with your quotient, yes...i got that

OpenStudy (anonymous):

x^5-29x^4+2x^3-58x^2+x-29

OpenStudy (anonymous):

so x^2+1=0 x^2=-1 x= +/- i, i multiplicity 2, -i multiplicity 2, 29 are the zeros

OpenStudy (anonymous):

awesome! thats what i got.

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