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alexamoyaaax3

Algebra II What are the possible number of positive, negative, and complex zeros of f(x) = -3x4 - 5x3 - x2 - 8x + 4 ? Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 Positive: 1; Negative: 3 or 1; Complex: 2 or 0 Positive: 3 or 1; Negative: 1; Complex: 2 or 0 Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

  • one year ago
  • one year ago

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  1. alexamoyaaax3
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    PLEEEEEAAAAAAAAASSSSEEEEE HEEEELLLLPPPPP ! & EXPLAIN :C

    • one year ago
  2. LilHefner3
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    r u n ca

    • one year ago
  3. some_someone
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    f(x)=-3x^(4)-5x^(3)-x^(2)-8x+4 If a polynomial function has integer coefficients, then every rational zero will have the form (p)/(q) where p is a factor of the constant and q is a factor of the leading coefficient. p=4_q=3 Find every combination of \(p)/(q). These are the possible roots of the polynomial function. \1,\(1)/(3),\2,\(2)/(3),\4,\(4)/(3)

    • one year ago
  4. some_someone
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    I would say; Positive: 1; Negative: 3 or 1; Complex: 2 or 0

    • one year ago
  5. some_someone
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    f(x)=-3x^(4)-5x^(3)-x^(2)-8x+4 Since there are 1 sign change from the highest order term to the lowest, there is at most 1 positive root (Descartes' Rule of Signs). There is 1 positive root. To find the possible number of negative roots, replace x with -x and repeat the sign comparison. f(-x)=-3(-x)^(4)-5(-x)^(3)-(-x)^(2)-8(-x)+4 Simplify the polynomial. f(-x)=-3x^(4)+5x^(3)-x^(2)+8x+4 Since there are 3 sign changes from the highest order term to the lowest, there are at most 3 negative roots (Descartes' Rule of Signs). There are 3 or 1 possible negative roots.

    • one year ago
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