SOLUTION: Use Descartes Rule of Signs to state the number of possible positive real zeros and the number of negative real zeros of P(x)=3x^4 - 6x^3 + 2x^2 - 7x

SOLUTION: Use Descartes Rule of Signs to state the number of possible positive real zeros and the number of negative real zeros of P(x)=3x^4 - 6x^3 + 2x^2 - 7x - 4 Please show the work s

Question 887075: Use Descartes Rule of Signs to state the number of possible positive real zeros and the number of negative real zeros of P(x)=3x^4 - 6x^3 + 2x^2 - 7x - 4
Please show the work so I can see how the problem was worked out. Thanks
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Use Descartes Rule of Signs to state the number of possible positive real zeros and the number of negative real zeros of P(x)=3x^4 - 6x^3 + 2x^2 - 7x - 4
Please show the work so I can see how the problem was worked out. Thanks
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P(x)=3x^4 - 6x^3 + 2x^2 - 7x - 4
# of changes of sign in P(x) = 3
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Using Descarte's Rule::
# of Real Roots <= # of sign changes in P(x) - 0 or 2
For P(x) that means # of Real Roots in P(x) is 3 or 1
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P(-x) = 3x^4 + 6x^3 + 2x^2 + 7x - 4
# of sign changes in P(-x) = 1
Using Descarte's Rule::
# of Real Roots <= # of sign changes in P(-x) - 0
For P(-x) that means # of Real Roots in P(x) is 1
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Conclusion for your problem::
# of Real Roots is 1
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For additional narrative and examples using Descarte's Rule
Google "Descartes Rule",
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Cheers,
Stan H.
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