The polynomial $p(x)=x^4+a x^3+b x^2+c x+d$ has exactly one real number $r$ such that $p(r)=0$. Show that if $a, b, c, d$ are rational, $r$ is also rational.
As it's a 4 degree polynomial with only one real root $r$, we know it's a double root. Hence $p(x)= (x-r)^2 q_1(x)$. Now $p'(x)$ also has $r$ as root. Hence $p'(x)= 2(x-r) q_2(x)$. Here $q_2(x)$ is a quadratic polynomial. Now I'm stuck here, how do I proceed
The possibilities are either (i) $r$ is a double root and other two complex roots are of multiplicity one, or (ii) $r$ is of multiplicity four. In the latter case, $p(x)=(x-r)^4$ and $r=-a/4$ is rational. In the former case, the GCD $d(x)=(p(x),p'(x))$ is proportional to $x-r$ and can be computed inside $\mathbb{Q}[x]$ by Euclid's algorithm. So $r$ is a root of a rational polynomial of degree one, hence rational.
We have that $p(x) = (x-r)^2q(x)$, where $q(x)$ does not have real roots. Let's look at minimal polynomial of $r$, denote it as $m(x)$. Note that degree of $m(x)$ not more than $3$, because $r$ is the root of $p'(x).$ If it has degree $1$ then $r$ is rational. It can't have degree $2$ because $m(x)|p(x)$ and in this case $m(x)$ has two real roots. Then we get that $m(x) = \frac{p'(x)}{4}.$ But it is easy to check that $p(x) \neq p'(x) q(x)$ if $p(x) \neq (x-r)^4$. (Each root of $p'(x)$ must be the root of $p(x)$ with larger multiplicity.)
The polynomial $p(x)$ must have $r$ as a repeated root. Let's consider the factorization of $p$ over the rational numbers, and let $p_1$ be the irreducible factor of $p$ having $r$ as a root. Since $r$ is a repeated root of $p$, the factor $p_1$ must likewise occur multiple times in the factorization of $p$. Also the degree of $p_1$ is obviously at most $2$. So over the rationals we have one of the following three situations: (i) $p=p_1^4$ with $p_1$ of degree $1$; (ii) $p=p_1^2p_2$ with $p_1$ of degree $1$ and $p_2$ irreducible of degree $2$ (and with non-real roots); or (iii) $p=p_1^2$ with $p_1$ irreducible of degree $2$.
If $p_1$ has degree $1$ then $r$ is obviously rational, so we may assume it has degree $2$. But if a polynomial of degree $2$ with rational coefficients has precisely one real root, then that root must be rational (because the discriminant must then be zero). But this means that $p_1$ was not irreducible, which is a contradiction, showing that case (iii) can't occur in the first place.
