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use the given zero to find the remaining zeros of the fuction...f(x)=x to the 5power-6x4power+7x3power-10x2power zero:1+i

  • math -

    type it this way:

    f(x) = x^5 - 6x^4 + 7x^3 - 10x^2

    since complex number zeros always come in conjugate pairs, there is another zero of 1-i
    so (x -(1+i)) and (1 - (1-i)) are factors
    expanded gives us:
    (x-1-i)(x-1+i)
    = x^2 - x + ix - x + 1 - i - ix + i - i^2 , but i^2 = -1
    = x^2 - 2x + 2

    so we have
    f(x) = x^2(x^3 - 6x^2 + 7x-10)
    but we know that (x^2 - 2x + 2) must be a factor of
    (x^3 - 6x^2 + 7x - 10)

    that is:
    ( x^2 - 2x + 2)(......) = x^3 - 6x^2 + 7x - 10
    that would leave a binomial factor of (x - 5)

    so the zeros are
    x = 0 , 5 , 1+i , 1-i

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