anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
jst check for those value whose value for thoce given equation gives "0".& this shows that root is real .
anonymous
  • anonymous
I've already tried some
anonymous
  • anonymous
@ash2326 @Mertsj Any ideas?

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

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