pre-cal

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use the given zero to find the remaining zeros of the function.

h(x)=x^4-12x^3+36x^2+68x-525
zero:4-3i

  • pre-cal -

    complex zeros always come in conjugate pairs
    so if 4-3i is a root, so is 4+3i

    so (x - 4 -3i)(x - 4 + 3i) will be a factor
    = x^2 - 4x + 3ix - 4x + 16 - 12i - 3ix + 12i - 9i^2)
    = x^2 - 8x + 25

    Using long algebraic division
    (x^4 - 12x^3 + 36x^2 + 68x - 525) / (x^2 - 8x + 25)
    = x^2 - 4x - 21

    so for x^2 - 4x - 21 = 0
    (x-7)(x+3) = 0
    x = 7 or x = -3

    roots are :
    -3 , 7 , 4+3i , 4-3i
    =

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