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11 November, 14:38

Find the zero of f (x) = x^5-12x^2+32x

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  1. 11 November, 15:45
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    The zeroes of f (x) = x^5-12x^2+32x can be found by factoring,

    f (x) = x^5-12x^2+32x = (x-8) (x-4)

    By the zero product theorem, (x-8) = 0 or (x-4) = 0 which means

    x=8 or x=4.

    So the zeroes of f (x) are S={4,8}
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