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29 October, 22:31

Find all unique zeros of the function fix) 8x - 64x + 160x - 128-

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  1. 29 October, 23:13
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    Answer: Lets divide by 8. My god, those coefficients are large.

    x^3 - 8x^2 + 20x - 16 = 0

    Possible integral roots are: + 1, - 1, + 2, - 2, + 4, - 4, + 8, - 8

    If x = 2, f (x) = 8 - 32 + 40 - 16 = 0

    Hence, x-2 is a factor.

    Use synthetic division to get: x^2 - 6x + 8 = (x-4) (x-2).

    Therefore, there are two integral roots: + 2 and + 4.
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