SOLUTION: o NEED TO WRITE A POLYNOMIAL FUNCTION OF LEAST DEGREE THAT HAS REAL COEFFICIENTS, THE GIVEN ZERO, AND LEADING COEFFICIENT OF 1. First problem gave me the zeros of 2, -2 and -6i

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: o NEED TO WRITE A POLYNOMIAL FUNCTION OF LEAST DEGREE THAT HAS REAL COEFFICIENTS, THE GIVEN ZERO, AND LEADING COEFFICIENT OF 1. First problem gave me the zeros of 2, -2 and -6i       Log On


   

Question 377607: o NEED TO WRITE A POLYNOMIAL FUNCTION OF LEAST DEGREE THAT HAS REAL COEFFICIENTS, THE GIVEN ZERO, AND LEADING COEFFICIENT OF 1.
First problem gave me the zeros of 2, -2 and -6i
Therefore to start:
(X-2)(X+2)(X+6i)
I multiplied (X-2)(X+2) and got (x2 - 4)
next I multiplied (x2 - 4)(X+6i) using the FOIL method
x3 + 6x2i - 4x - 24i
I know that i2 = -1 and that i = rad -1 but don't know how to deal with the "i" in this problem to make it a polynomial function.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
If you have a complex root, then it's complex conjugate is also a root since complex roots only occur as complex conjugate pairs.
f%28x%29=1%2A%28x-2%29%28x%2B2%29%28x-6i%29%28x%2B6i%29
f%28x%29=%28x%5E2-4%29%28x%5E2%2B36%29
Take it from there to complete the expansion.