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2 September, 10:02

Find a polynomial with degree 3; zeros - 3,3,5

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Answers (2)
  1. 2 September, 10:13
    0
    Step-by-step explanation:

    f (x) = ax^3+bx^2+cx+d

    f (-3) = -3a^3-3b^2-3c+d=0

    f (3) = 3a^3+3b^2+3c+d=0

    f (5) = 5a^3+5b^2+5c+d=0

    f (0) = 0a^3+0b^2+0c+d=2 < = any number you want your extremum at

    => d=2

    f (-3) = -3a^3-3b^2-3c+=-2

    f (3) = 3a^3+3b^2+3c+=-2

    f (5) = 5a^3+5b^2+5c+=-2

    solve for all variables (Gauß)

    a = 0.0266

    b = - 0.08

    c = - 0.66

    As Solution

    0.0266x^3-0.08x^2-0.66x+2 approximately
  2. 2 September, 10:30
    0
    (x^3) - (5x^2) - 9x+45

    Explanation:

    (x-3) (x+3) (x-5)

    (x^2-9) (x-5)

    (x^3) - (5x^2) - 9x+45
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