SOLUTION: Find a polynomial of degree 2 with real-number coefficients and zero 3-2i

Question 1059919: Find a polynomial of degree 2 with real-number coefficients and zero 3-2i
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The zeros are 3 +/- 2i, since complex roots come as conjugate
This is the result of -b+/-sqrt(b^2-4ac) all divided by 2a
Therefore, b=-6, because it will be divided by 2 at the end
b^2-4ac=36-4*c, and that has to equal -16. The sqrt (-16) is +/-4i, and dividing that by 2 will make 2i
Therefore, 4c=52, and c=13.
x^2-6x+13=0

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