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December 18, 2014

December 18, 2014

Posted by **april** on Monday, December 2, 2013 at 6:04pm.

- math -
**Reiny**, Monday, December 2, 2013 at 6:31pmtype 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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