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Pre cal

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I need help finding the zeros of a function algebraically. And here is one of them f(x)= x^3-4x^2-9x+36

  • Pre cal -

    try writing as:
    (x^3 - 9x) -(4x^2-36) =0

    x(x^2-9) -4 (x^2-9)=0

    (x^2-9)(x-4) = 0

    (x-3)(x+3)(x-4) = 0

    3 , -3 , 4

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