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March 25, 2017

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Find the real and complex zeros of the following function. Please show all of your work.


f(x) = x^3 + 2x^2 - 6x + 8

  • find real and complex 0 - ,

    f(±1) ≠ 0
    f(±2) ≠ 0
    but f(-4) = 0
    so x+4 is a factor

    I then used synthetic division to get
    (x+4)(x^2 - 2x + 2) = 0

    the real root is x = -4

    Use the quadratic formula to get the two complex roots of
    1+i and 1-i

  • find real and complex 0 - ,

    f(x)=x^3+4x^2-2x^2-8x+2x+8=(x^3+4x^2)-
    (2x^2+8x)+(2x+8)=x^2(x+4)-2x(x+4)+2(x+4)=
    (x+4)(x^2-2x+2)
    x+4=0 => x=-4
    or
    x^2-2x+2=0 => x=1+i or x=1-i

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