How to find a polynomial from a given root? I was asked to find a polynomial with integer coefficients from a given root/solution.
Lets say for example that the root is: $\sqrt{5} + \sqrt{7}$.  


*

*How do I go about finding a polynomial that has this number as a root?

*Is there a specific way of finding a polynomial with integer coefficients?


Any help would be appreciated. Thanks.

The simplest polynomial that has $r$ as a root is just $x-r$. But of course, that is not what you are dealing with here.
This is really a question about finding the minimal polynomial of $\alpha=\sqrt{5}+\sqrt{7}$ over $\mathbb{Q}$; such polynomials exist for any algebraic number, by definition of algebraic.
Galois Theory provides all the necessary tools to solve this problem. Basically:


*

*Consider the field $\mathbb{Q}(\sqrt{5},\sqrt{7})$; this field contains the number $\alpha=\sqrt{5}+\sqrt{7}$, so we can work there. Every element of this field can be written uniquely as
$$a + b\sqrt{5} + c\sqrt{7} + d\sqrt{35}$$
for some $a,b,c,d\in\mathbb{Q}$.

*The field $\mathbb{Q}(\sqrt{5},\sqrt{7})$ has four automorphisms (functions $f\colon\mathbb{Q}(\sqrt{5}+\sqrt{7})\to\mathbb{Q}(\sqrt{5},\sqrt{7})$ that are additive, $f(a+b)=f(a)+f(b)$, multiplicative, $f(ab)=f(a)f(b)$, and invertible), and these automorphisms fix elements of $\mathbb{Q}$ (that is, $f(q)=q$ for all $q\in\mathbb{Q}$).
These automorphism are the following:
$$\begin{align*}
\sigma_1\colon a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\longmapsto a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\qquad&\text{(the identity)};\\
\sigma_2\colon a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\longmapsto a-b\sqrt{5}+c\sqrt{7}-d\sqrt{35}\\
\sigma_3\colon a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\longmapsto a+b\sqrt{5}-c\sqrt{7}-d\sqrt{35}\\
\sigma_4\colon a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35} &\longmapsto a-b\sqrt{5}-c\sqrt{7}+d\sqrt{35} &&\text{(equal to }\sigma_3\circ\sigma_2\text{)}
\end{align*}$$
The maps are induced by the conjugation maps $\sqrt{5}\mapsto-\sqrt{5}$ and $\sqrt{7}\mapsto-\sqrt{7}$.
An important feature is that an element of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ lies in $\mathbb{Q}$ if and only if it is fixed by all four maps. 

*Suppose $p(x)$ is a polynomial with coefficients in $\mathbb{Q}$, and that $a\in\mathbb{Q}(\sqrt{5},\sqrt{7})$. If
$$p(x) = a_nx^n + \cdots + a_0$$
then
$$\sigma_i(p(a)) = \sigma_i(a_na^n+\cdots+a_0) = a_n\sigma_i(a)^n+\cdots+a_0 = p(\sigma_i(a)).$$
In particular, if $p(a)\in\mathbb{Q}$, then $p(a)=\sigma_i(p(a)) = p(\sigma_i(a))$ for $i=1,2,3,4$. 

*Now suppose you find a polynomial $p(x)$ with coefficients in $\mathbb{Q}$ that has $\sqrt{5}+\sqrt{7}$ as a root. Then $p(\sqrt{5}+\sqrt{7})=0$, so by 3 above, it must also be true that $p(\sigma_i(\sqrt{5}+\sqrt{7}))=0$ for $i=1,2,3,4$. That means that $p(x)$ must also have $\sqrt{5}-\sqrt{7}$, $-\sqrt{5}+\sqrt{7}$, and $-\sqrt{5}-\sqrt{7}$ as roots. By unique factorization, we conclude that $p(x)$ must be divisible by
$$\left(x - (\sqrt{5}+\sqrt{7})\right)\left(x - (-\sqrt{5}+\sqrt{7})\right)\left(x - (\sqrt{5}-\sqrt{7})\right)\left(x - (-\sqrt{5}-\sqrt{7})\right).$$
That is, any polynomial with coefficients in $\mathbb{Q}$ that has $\sqrt{5}+\sqrt{7}$ as a root must be a multiple of this product.
But if you multiply out this product, you will discover that this polynomial already has coefficients in $\mathbb{Q}$.
Once you have a polynomial with coefficients in $\mathbb{Q}$ that has $\sqrt{5}+\sqrt{7}$ as a root, you can get a polynomial with coefficients in $\mathbb{Z}$ by simply clearing denominators.
