SOLUTION: Apply Descartes' Rule of Signs to f(x)=5x^3 - x^2 - 1 to determine the number of positive and negative zeros.
a) # of positive zeros:
b) # of negative zeros:
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a) # of positive zeros:
b) # of negative zeros:
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Question 949929: Apply Descartes' Rule of Signs to f(x)=5x^3 - x^2 - 1 to determine the number of positive and negative zeros.
a) # of positive zeros:
b) # of negative zeros: Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! here's a tutorial on descarted rule of signs. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut38_zero1.htm
f(x) = 5x^3 - x^2 - 1
signs are + - -
there is one sign change so there will be one positive real root.
f(-x) = 5(-x)^3 - (-x)^2 - 1
this simplifies to:
f(-x) = -5x^3 - x^2 - 1
signs are - - -
there are no sign changes.
there are no negative real roots.
since the degree of the equation is 3 and there is only 1 positive real root, then the number of complex roots must be equal to 2.
the graph of your equation is shown below:
there is one positive root.
since the degree of the equation is 3, there must be 3 roots.
2 of them must be complex.
1 of them is positive.
none of them are negative.
real roots cross the x-axis.
complex roots don't.
