SOLUTION: Find a polynomial of lowest degree that has real coefficients, a leading coefficient of 1, and with 3 and 2+i as two of its zeros. I got as far as: f(x)= (x-3)(x-(2+i))(x-(2-i)

Algebra ->  Rational-functions -> SOLUTION: Find a polynomial of lowest degree that has real coefficients, a leading coefficient of 1, and with 3 and 2+i as two of its zeros. I got as far as: f(x)= (x-3)(x-(2+i))(x-(2-i)      Log On


   

Question 604244: Find a polynomial of lowest degree that has real coefficients, a leading coefficient of 1, and with 3 and 2+i as two of its zeros.
I got as far as:
f(x)= (x-3)(x-(2+i))(x-(2-i)), but then I am completely stuck.
I appreciate the help, thank you!
Answer by flame8855(424) About Me  (Show Source):
You can put this solution on YOUR website!
(x-3)(x-(2+i))(x-(2-i))= (x-3) ( (x-2)^2+1))= (x-3) ( x^2-4x+5)
= x^3 -7x^2+7x-15