SOLUTION: find an nth degree polynomial funcition with real coefficients satisfying the given conditions. n=3; 6 and -5+2i are zeroes; f(2)=-636 I understand that x=6 and x=-5+2i but what

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: find an nth degree polynomial funcition with real coefficients satisfying the given conditions. n=3; 6 and -5+2i are zeroes; f(2)=-636 I understand that x=6 and x=-5+2i but what      Log On


   

Question 743717: find an nth degree polynomial funcition with real coefficients satisfying the given conditions. n=3; 6 and -5+2i are zeroes; f(2)=-636
I understand that x=6 and x=-5+2i but what do I do about the imaginary number in -5+2i?
Answer by josgarithmetic(39627) About Me  (Show Source):
You can put this solution on YOUR website!
n=3, and you have TWO imaginary roots. The conjugate of the given imaginary root is also a root.

Your function is like this: roots 6, -5+2i, and -5-2i,
a%2A%28x-6%29%28x-%285-2i%29%29%28x-%285%2B2i%29%29, be sure you understand how the expression was chosen.
If p is a root of a polynomial function then the function contains a binomial factor of (x-p). If -p is a root of a polynomial function then the function contains a binomial factor of %28x-%28-p%29%29=%28x%2Bp%29. This means that for a complex root you must watch the signs very carefully when you form a binomial factor.

You want to multiply the two complex factors in order to later have computational convenience....,
...
If my own steps doing so are done correctly, then that would be x%5E2-10x%2B29 ?
So your function is like a%2A%28x-6%29%28x%5E2-10x%2B29%29.

You still want to figure what is the factor, a, through using f%282%29=-636.