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anonymous
 4 years ago
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!
anonymous
 4 years ago
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!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0jst check for those value whose value for thoce given equation gives "0".& this shows that root is real .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I've already tried some

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ash2326 @Mertsj Any ideas?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So it's either a or d.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1One integer and two rationals.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did you get that?
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