(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
Use the given zero to find the remaining zeros of the function:
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?
1 answer