SOLUTION: Please help me with this equation-Thank You! Find a polynomial equation with real coefficients that has the given roots. 3, -9, 3+2i

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Question 135278: Please help me with this equation-Thank You!
Find a polynomial equation with real coefficients that has the given roots.
3, -9, 3+2i

Found 2 solutions by scott8148, solver91311:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
if r is a root, then x-r is a factor

complex roots (3+2i) occur in conjugate pairs, so 3-2i is also a root

(x-3)(x+9)(x-3-2i)(x-3+2i)=0

multiply the factors to find the equation
__ HINT: start with the complex factors, it will be less "messy"

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Two things:
First, if a polynomial equation has a complex number root (a%2Bbi), then it also has the conjugate of that complex number as a root (a-bi). That means that, although you were only given three numbers as roots, there are actually four, namely:
3, -9, 3%2B2i, and 3-2i

Second, a polynomial equation has a root a if and only if %28x-a%29 is a factor of the polynomial.

So if the desired polynomial is P%28x%29, then P%28x%29=%28x-3%29%28x%2B9%29%28x-%283%2B2i%29%29%28x-%283-2i%29%29. All you need to do now is multiply the factors and collect like terms. Hint: Remember when you are working it out that i%5E2=-1, so 2i%28-2i%29=4 not -4.