SOLUTION: What are possible number of positive ,negative and imaginary zeros of P(x) . P(x)= 2x^5 + x^3

SOLUTION: What are possible number of positive ,negative and imaginary zeros of P(x) . P(x)= 2x^5 + x^3 - 2x^2 + 3x -5

Question 1180083: What are possible number of positive ,negative and imaginary zeros of P(x) .
P(x)= 2x^5 + x^3 - 2x^2 + 3x -5
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Use Descartes' rule of signs.

In P(x), there are 3 sign changes; the possible numbers of positive real roots are 3 or 1.

In P(-x), there are 0 sign changes; the number of negative real roots is 0.

ANSWER: There are two possibilities:
(a) 3 positive real roots and one pair of imaginary roots; or
(b) 1 positive real root and two pairs of imaginary roots

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