Find a monic polynomial of degree 4, in x, with rational coefficients such that sqrt(2) + sqrt(5) is a root of the polynomial.
+287 Let $k = \sqrt2+\sqrt5$. We first compute $k^2$, $k^3$, $k^4$. They are
\begin{eqnarray*}
k^2 &=& 2\sqrt{10}+7 \\
k^3 &=& 11\sqrt5+17\sqrt2 \\
k^4 &=& 28\sqrt{10}+89.
\end{eqnarray*}
Let the polynomial be $P(x)=x^4+bx^3+cx^2+dx+e$. We have
\begin{eqnarray*}
P(k) &=& k^4 + bk^3+ck^2+dk+e \\
&=& (28\sqrt{10}+89) + (11\sqrt5+17\sqrt2)b + (2\sqrt{10}+7)c + (\sqrt2+\sqrt5)d + e \\
&=& (28 + 2c) \sqrt{10} + (11b + d)\sqrt5 + (17b + d)\sqrt2 + (89 + 7c + e).
\end{eqnarray*}
The only way for $P(k)$ to be 0 is that
\begin{eqnarray*}
28 + 2c &=& 0 \\
11b + d &=& 0 \\
17b + d &=& 0 \\
89 + 7c + e &=& 0
\end{eqnarray*}
Solving the above system of equations gives $b=0$, $c=-14$, $d=0$, and $e=9$.
Hence, the disired polynomial is $P(x) = x^4 - 14x^2 + 9$.
