What are the possible numbers of positive real, negative real, and complex zeros of f(x) = 6×3 − 3×2 + 5x + 9?

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What are the possible numbers of positive real, negative real, and complex zeros of f(x) = 6×3 − 3×2 + 5x + 9?

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Xavia 3 months 2021-09-01T07:16:01+00:00 1 Answers 0 views 0

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    2021-09-01T07:17:23+00:00

    Answer:

    Positive Roots: 2 or 0

    Negative Roots: 1

    Step-by-step explanation:

      f (x) = 6×3 − 3×2 + 5x + 9

    To find the possible number of positive roots, look at the signs on the coefficients and count the number of times the signs on the coefficients change from positive to negative or negative to positive.

      f (x) = 6×3 − 3×2 + 5x + 9

    Since there are 2 sign changes from the highest order term to the lowest, there are at most 2 positive roots (Descartes’ Rule of Signs). The other possible numbers of positive roots are found by subtracting off pairs of roots (2 − 2).

    Positive Roots: 2 or 0

    To find the possible number of negative r oots, replace x with −x and repeat the sign comparison.

      f (−x) = 6(−x)3 − 3(−x)2 + 5 (−x) + 9

    Simplify each term.

      Apply the product rule to −x.

         f (−x) = 6 ((−1)3x^3) − 3(−x)^2 + 5 (−x) + 9

       Raise −1 to the power of 3.

         f (−x) = 6 (−x^3) − 3(−x)^2 + 5 (−x) + 9

       Multiply −1 by 6.

         f (−x) = −6x^3 − 3(−x)^2 + 5 (−x) + 9

       Apply the product rule to −x.

         f (−x) = −6x^3 − 3 ((−1)^2(x^2)) + 5 (−x) + 9

    Raise −1 to the power of 2.

         f (−x) = −6x^3 − 3 (1x^2) + 5 (−x) + 9

    Since there is 1 sign change from the highest order t erm to the lowest, there is at most 1 negative

    root (Descartes’ Rule of Signs). Negative Roots: 1

    The possible number of positive roots is 2 or 0, and the possible number of negative roots is 1. Positive Roots: 2 or 0

    Negative Roots: 1

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