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20 September, 03:34

What are the real zeros of f (x) = x^3 + 2x^2-5x-6

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  1. 20 September, 06:21
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    First of all Horner

    First we try with - 3

    1. 2. - 5. - 6.

    -3. 3. 6

    1. - 1. - 2. 0

    To verify: - 3^3+2*-3^2+15-6=0

    Now we have this equation

    (X+3) * (x^2 - x - 2) = 0

    I dont know If you know sum and product but that is the easiest way so sum = 1 product = 2 so 2 and - 1 are the answers.

    You can do this too by this way : b^2-4ac = D

    X = (-b (+or-) d^1/2) / 2a

    So all the zeros

    2, - 1 and - 3

    If you want more explanation, send me a comment x
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