Question 203583
i need to find the positive root, negative roots, and rational roots. 
Given x^3-4x^2+2x+1=0 

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Since the coefficients add up to zero, x=1 is a root
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Using synthetic division you get the following:
1)....1....-4....2....1
........1.....-3...-1..|..0

Quotient: x^2 - 3x -1
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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)
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x = 1 is a rational root and is positive
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x = (3/2)+(1/2)sqrt(13) is a positive root but not rational
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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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