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28 October, 23:55

Find algebraically the zeros for p (x) = x^3+x^2-4x-4

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  1. 29 October, 00:25
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    One zero is - 1

    because p (-1) = (-1) ^3 + (-1) ^2 - 4 (-1) - 4 = 0

    So one factor of p (x) is x + 1

    Dividing p (x) by x + 1 gives the quotient

    x^2 - 4

    x - 2) (x + 2) = 0

    x = 2, - 2

    The zeroes are - 2,-1 and 2
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