SOLUTION: I need help with this problem, please help me: i need to find the positive root, negative roots, and rational roots. Given x^3-4x^2+2x+1=0 i got for the positive root=2

Question 203583: I need help with this problem, please help me:
i need to find the positive root, negative roots, and rational roots.
Given x^3-4x^2+2x+1=0

i got for the positive root=2
negative root=1
and im having trouble finding the rational roots.
thank you in advance!

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
i need to find the positive root, negative roots, and rational roots.
Given x^3-4x^2+2x+1=0
---------------------------
Since the coefficients add up to zero, x=1 is a root
---------------
Using synthetic division you get the following:
1)....1....-4....2....1
........1.....-3...-1..|..0
Quotient: x^2 - 3x -1
-----------------
Use the quadratic formula to get:
x = [3 +- sqrt(9-4*1*-1)]/2
x = [3 +- sqrt(13)]/2
x = (3/2) + (1/2)sqrt(13) or x = (3/2)-(1/2)sqrt(13)
--------------------
x = 1 is a rational root and is positive
---
x = (3/2)+(1/2)sqrt(13) is a positive root but not rational
--------
x = (3/2)-(1/2)sqrt(13) is a negative root but not rational.
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Cheers,
Stan H.
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