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What can we afirm about p(x)=x^5 - 5x^3 + 4x^2 - 3x - 2? a) p(x) has only 3 real roots (1 integer and 2 rationals) b)p(x) has only one real root, which is also integer c) x=2 isn't a root of p(x) d)p(x) has only real roots (1 integer, 2 rationals and 2 irrationals) So, c is wrong because I divided p(x) by x - 2 and I got x^4 + 2x^3 - x^2 + 2x + 1. Problem is I have no idea how to find the other roots! Help!

Mathematics
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jst check for those value whose value for thoce given equation gives "0".& this shows that root is real .
I've already tried some
@ash2326 @Mertsj Any ideas?

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Other answers:

2 is one rational zero
So it's either a or d.
It has three real zeros
One integer and two rationals.
A
how did you get that?
i graphed it on Wolfram
Clever. Thx
yw

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