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Working on question 1 here

Find a polynomial with integer coefficients that has the following zeros: -1/3, 2, 3+i

Multiplying (3x+1) (x-2) (x-3-i) produces a polynomial with i scattered throughout the terms. Not the right answer.

Then I thought maybe the root 3+i implied another root 3-i.

That didn't produce the right answer either.

What am I missing here?

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'Then I thought maybe the root $3+i$ implied another root $3-i$'. You thought well. What did you do exactly to reach the conclusion that it didn't work? – Git Gud Dec 11 '13 at 23:20
You are in the right track. The complex roots always come as conjugated. Thus the equation will actually have 4 roots. Then you can rewrite into $(x-a)(x-b)(x-c)(x-d)=0$ form, and you will have your answer. – LorenMt Dec 11 '13 at 23:21
Is your next factor $x+3-i$ instead of $x-3+i$ perhaps? – Ross Millikan Dec 11 '13 at 23:26

Hint: a polynomial with integer coefficients that has the root $3+i$ must have also the root $3-i$. However

$$ (3x+1) (x-2) (x-3-i) $$

doesn't have the root $3-i$. How can you make from this a polynomial that has also the desired root?

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Alpha agrees with the solution on the page you link to $ (3x+1)(x-3-i)(x-3+i)(x-2)=3 x^4-23 x^3+58 x^2-38 x-20$

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