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chetan
chetan
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Posted on : 01:17:13 PM on July 21, 2009

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!

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Vasu sharma
  Vasu
Posted on : 01:59:18 PM on July 21, 2009
i need to find the positive root, negative roots, and rational roots.
Given x^{3}-4 x^{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+-\frac{ \sqrt 9-4\times 1\times-1} {2}
x= 3+-\frac{ \sqrt 13} {2}
x= \frac{3}{2}+\frac{ 1} {2}\sqrt{13} \text or x= \frac{3}{2}-\frac{ 1}{2}\sqrt{13}
--------------------
x = 1 is a rational root and is positive
---
x= \frac{3}{2}+\frac{ 1}{2}\sqrt{13} is a positive root but not rational
--------
x= \frac{3}{2}-\frac{ 1}{2}\sqrt{13} is a negative root but not rational.



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