Question

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!

Answer 1

jst check for those value whose value for thoce given equation gives "0".& this shows that root is real .

Answer 2

I've already tried some

Answer 3

@ash2326 @Mertsj Any ideas?

Answer 4

2 is one rational zero

Answer 5

So it's either a or d.

Answer 6

It has three real zeros

Answer 7

One integer and two rationals.

Answer 8

A

Answer 9

how did you get that?

Answer 10

i graphed it on Wolfram

Answer 11

Clever. Thx

Answer 12

yw