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1) Suppose $P(x)$ is a polynomial of smallest possible degree such that:

$\bullet$ $P(x)$ has rational coefficients

$\bullet$ $P(-3) = P(\sqrt 7) = P(1-\sqrt 6) = 0$

$\bullet$ $P(-1) = 8$

Determine the value of $P(0)$.

 This is a degree of $3$, correct?

2) Find a monic quartic polynomial $f(x)$ with rational coefficients whose >roots include $x=3-i\sqrt[4]2$. Give your answer in expanded form.

 How would I do this?
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    Answer the first question for each one individually to get the idea. Here's a start: If $P$ has rational coefficients and $P(\sqrt{7})=0$, then $P(x)=x-\sqrt{7}$ is the polynomial of smallest possible degree but it does not have rational coefficients. Hence, you need $P(x)=x^2-7$. – 2017-02-19
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    Hint for #2: $\;(x-3)^2=-\sqrt{2}\,$ which leaves only one radical to eliminate. – 2017-02-19

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