SOLUTION: I have three zeros 3-3i,3+3i, and 3. I am having trouble factoring. I am missing something in the factoring that will get me the right equation. This is what I have so far (x-(3-3i

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I have three zeros 3-3i,3+3i, and 3. I am having trouble factoring. I am missing something in the factoring that will get me the right equation. This is what I have so far (x-(3-3i      Log On


   

Question 1123854: I have three zeros 3-3i,3+3i, and 3. I am having trouble factoring. I am missing something in the factoring that will get me the right equation. This is what I have so far (x-(3-3i))(x-(3+3i))(x-3). (x-(3-3i))(3+3i)) should factor out to x^2-6x+10 I keep getting x^2-2x+18. The final equation should be x^3-9x^2+28x-30. Could you please assist me with this by working out the factoring to see where I am going wrong.
Thanks
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

zeros:+x%5B1%5D=3-3i,x%5B2%5D=3%2B3i, and x%5B3%5D=3
to find equation, use product of factors
%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29...substitute given zeros
%28x-%283-3i%29%29%28x-%283%2B3i%29%29%28x-3%29
%28x-3%2B3i%29%28x-3-3i%29%28x-3%29

%28x%5E2-6x%2B9%2B9%2A1%29%28x-3%29
%28x%5E2-6x%2B9%2B9%29%28x-3%29
%28x%5E2-6x%2B18%29%28x-3%29
x%5E3-6x%5E2%2B18x+-3x%5E2%2B18x-54
x%5E3-9x%5E2%2B36x-54