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Precalc

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Given g(x) = (x + 5)(x2 - 6x + 13), find the one real zero of g(x):

I think it's -5

Find the two complex zeros of g(x):

  • Precalc -

    *(x + 5)(x^2 - 6x + 13)

  • Precalc -

    correct for real root x = -5

    for complex:
    x^2 - 6x + 13 = 0
    x =(6 ± √-16)/2
    = (6 ± 4i)/2
    = 3 ± 2i

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