SOLUTION: write a polynomial function with rational coefficients so that it has roots (-2, 1+3i )

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Question 709832: write a polynomial function with rational coefficients so that it has roots (-2, 1+3i )
Answer by josgarithmetic(39628) About Me  (Show Source):
You can put this solution on YOUR website!
You seem to be asking for polynomial function with roots: -2,and 1+3i, AND 1-3i. Complex roots come as conjugate pairs.

-2 is a root:
((-2)-(k))=0
-2-k=0
-k=2
k=-2
The binomial factor is (x-(-2))=%28x%2B2%29

1+3i is a root:
(1+3i-k)=0
1+3i-k=0
-k=-1-3i
The binomial factor is (x-(-1-3i))
or equal to+%28x%2B%281%2B3i%29%29

1-3i is a root:
(1-3i-k)=0
-k=-1+3i
k=1-3i
The binomial factor is %28x-%281-3i%29%29

A satisfying function can come from %28x%2B2%29%28x%2B%281%2B3i%29%29%28x-%281-3i%29%29

Do the multiplications, at least for the factors containing the complex numbers:
... The finished polynomial function may be:
highlight%28f%28x%29=%28x%2B2%29%28x%5E2-2x%2B10%29%29 and you could do the rest of the multiplying if needed.