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**Neha****Member**- Registered: 2006-10-11
- Posts: 173

Apply Descartes' Rule of Signs to determine the possible numbers of positive roots of:

x^4 - x^3 - 3x^2 - x + 2 = 0

but not sure what i did is right....correct me if i did it wrong

f(x) = +x^4 - x^3 - 3x^2 - x + 2

f(-x) = -x^4 - (-x)^3 - 3(-x)^2 - (-x) + 2

= -x^4 + x^3 + 3x^2 + x + 2

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**er.neerajsrivastava****Member**- Registered: 2006-12-27
- Posts: 9

no of possitive roots = the no of times sign is changed in the eq f(x)

and no of negative roots = the no of times sign changes in eq f(-x).

so for the given eq. +ve root = 2

and -ve root =

f(-x)= x^4 + x^3 -3x^2 +x +2=0

so -ve root = 2

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Hm, Interesting, I didn't know that before.

IPBLE: Increasing Performance By Lowering Expectations.

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