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The given zeros are $-5$, $-5$, $1+ \sqrt{3} i$.

I know how to get the polynomials for $-5$ but I have no idea how to in $1+ \sqrt{3} i$.

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    Are you looking for a polynomial with real or complex roots? In the first case, you'll have a polynomial of degree 4, in the second case of degree 3. – 2017-02-27
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    Must the coeficients be real? Rational. The simplest is $P(x) = (x + 5)(x+5)(x+\sqrt{3}i)$. I'm assuming five is a double root. – 2017-02-27
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    The polynomial $\,(x+5)(x-1-i\sqrt{3})=x^2+(4-i\sqrt{3})\,x -5 -5i\sqrt{3}\,$ has both $\,-5\,$ and $\,1+i\sqrt{3}\,$ as zeros. If this is not the answer you wanted, think at what's missing from the question you asked. – 2017-02-27

2 Answers 2

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Hint: If $r$ is a root of a polynomial $P(x)$, then $x-r$ divides $P(x)$, regardless of whether $r$ is real or not. This also applies to double roots (e.g. the $-5$ given in your example).

Given this information, can you figure out what the polynomial should be?

Edit: There is another property that if a polynomial with integer coefficients has a complex root $a+bi$ (for reals $a$ and $b$), then $a-bi$ must also be a root of that polynomial. This information may also be helpful.

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These roots describe a quadratic of the form: $f(x) = x^{4} + bx^{3} + cx^{2} + dx + e =0$

with roots $\alpha = -5, \beta = 1+ \sqrt{3}i, \gamma = 1-\sqrt{3}i$

Then $f(x) = (x-\alpha)^{2}(x-\beta)(x-\gamma) = 0$

Comparing coeffiicients gives us:

$\alpha^{2}\beta\gamma = e$

$-(2\beta\alpha\gamma + \beta\alpha^{2} + \gamma\alpha^{2}) = d$

$\alpha^{2} + \beta\gamma + 2\alpha\beta + 2\alpha\gamma = c$

$-(2\alpha + \beta + \gamma) = b$

Note that $\beta\gamma = (1+\sqrt{3}i)(1-\sqrt{3}i) = 4$

and $\beta + \gamma = 2$

Then we have $e=100, d =-10$

$c = 9$

$b = 8$

Then $f(x) = x^{4} +8x^{3} +9x^{2} -10x + 100$

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    Why the downvote, is there something wrong in my answer? – 2017-02-27