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16 October, 02:34

Find all zeros of

f (x) = x^3 + 3x^2 - 3x - 1

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  1. 16 October, 05:30
    0
    F (1) = 1^3 + 3 (1) ^1 - 3 - 1 = 1 + 3 - 3 - 1 = 0

    so one zero is x = 1, and x-1 is a factor

    dividing f (x) by x - 1 gives x^2 + 4x + 1

    x = [-4 + / - sqrt (16 - 4) ] / 2

    x = - 0.27, - 3.73

    zeroes are 1, - 0.27 and - 3.73 Answer
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