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July 29, 2014

July 29, 2014

Posted by **James** on Wednesday, October 10, 2012 at 12:39am.

h(x)=x^4-15x^3+55x^2+155x-1476 Zero:5-4i

I multiplied [x-(5-4i)][x-(5+4i)] and got x^2-10x+9.

The example I have says to divide that by h and get a second quadratic equation that is also a factor of h. I have to use that equation to find the remaining zeros. How do I find the second equation?

- Pre-calc -
**Reiny**, Wednesday, October 10, 2012 at 12:54am(x-(5-4i))(x-(5+4i))

= (x-5+4i)(x-5 - 4i)

= x^2 - 5x - 4ix - 5x + 25 + 20i + 4ix -20i - 16i^2

= x^2 - 10x + 41

now divide x^4-15x^3+55x^2+155x-1476 by x^2 - 10x + 41 to get

x^2 - 5x - 36

(Google " long algebraic division" if you don't know how to do that)

now solving this quadratic:

x^2 - 5x - 36 = 0

(x-9)(x+4) = 0

x = 9, or x = -4

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