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14 July, 12:03

Find zeros for f (x) = x^3 - 3x^2 - x + 3

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  1. 14 July, 12:29
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    f (x) = x³ - 3x² - x + 3

    0 = x - 3x² - x + 3

    0 = x² (x) - x² (1) - 1 (x) + 1 (3)

    0 = x² (x - 3) - 1 (x - 3)

    0 = (x² - 1) (x - 3)

    0 = (x² - x + x - 1) (x - 3)

    0 = (x (x) - x (1) + 1 (x) - 1 (1)) (x - 3)

    0 = (x (x - 1) + 1 (x - 1)) (x - 3)

    0 = (x + 1) (x - 1) (x - 3)

    0 = x + 1 U 0 = x - 1 U 0 = x - 3

    0 - 1 = x + 1 - 1 U 0 + 1 = x - 1 + 1 U 0 + 3 = x - 3 + 3

    -1 = x U 1 = x U 3 = x

    Solution Set: {-1, 1, 3}
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