i need all possible rational roots
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There are two changes in sign, so that implies at most two positive roots. If we change the signs of the odd powered terms we have only one change of sign, implying at most one negative root. Also, we have rational factors of 2 as 2 and 1 only. This gives us a maximum of one rational root, and it must be +2 or +1. It can't be the latter because 1-7+2 is non-zero and -1+7+2 is non-zero.

By trial we see that x=-2 is a root because -16+14+2=0. If we use synthetic division to divide by this root we get:

-2 | 2..0 -7...2

......2 -4..8..-2

......2 -4..1 | 0

2x^2-4x+1, so x=(4+sqrt(16-8))/4 (irrational roots)=(2+sqrt(2))/2 [1.7071 and 0.2929].

So the only rational root is x=-2, the other roots being real but irrational.

by Top Rated User (1.2m points)

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