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December 22, 2014

December 22, 2014

Posted by **ladybug** on Sunday, October 14, 2012 at 12:08am.

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

zero:4-3i

- pre-cal -
**Reiny**, Sunday, October 14, 2012 at 1:01amcomplex 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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