Problem: Let $n > k \ge 1$ be integers.
Let $P(x) \in \RR[x]$ be a polynomial of degree $n$
with no repeated roots and $P(0) \neq 0$.
Suppose that for any real numbers $a_0$, \dots, $a_k$
such that the polynomial $a_k x^k + \dots + a_1 x + a_0$ divides $P(x)$,
the product $a_0 a_1 \dots a_k$ is zero.
Prove that $P(x)$ has a nonreal root.

Solution: By considering any $k+1$ of the roots of $P$, we may as well assume WLOG that $n = k+1$.
Suppose that $P(x) = (x+r_1) \dots (x+r_n) \in \RR[x]$ has $P(0) \neq 0$.
Then the problem hypothesis is that each of the $n$ polynomials
(of degree $n-1$) given by
\begin{align*}
  P_1(x) &= (x+r_2)(x+r_3)(x+r_4) \dots (x+r_n) \\
  P_2(x) &= (x+r_1)(x+r_3)(x+r_4) \dots (x+r_n) \\
  P_3(x) &= (x+r_1)(x+r_2)(x+r_4) \dots (x+r_n) \\
  &\vdotswithin= \\
  P_n(x) &= (x+r_1)(x+r_2)(x+r_3) \dots (x+r_{n-1})
\end{align*}
has at least one coefficient equal to zero.
(Explicitly, $P_i(x) = \frac{P(x)}{x+r_i}$.)
We'll prove that at least one $r_i$ is not real.

Obviously the leading and constant coefficients of each $P_i$ are nonzero,
and there are $n-2$ other coefficients to choose between.
So by pigeonhole principle, we may assume, say,
that $P_1$ and $P_2$ share the position of a zero coefficient, say
the $x^k$ one, for some $1 \le k < n-1$.

\begin{claim*}
  If $P_1$ and $P_2$ both have $x^k$ coefficient equal to zero, then the polynomial
  \[ Q(x) = (x+r_3)(x+r_4) \dots (x+r_n) \]
  has two consecutive zero coefficients, namely $b_k = b_{k-1} = 0$.
\end{claim*}
\begin{proof}
  Invoking Vieta formulas, suppose that
  \[ Q(x) = x^{n-2} + b_{n-3} x^{n-3} + \dots + b_0. \]
  (And let $b_{n-2} = 1$.)
  Then the fact that the $x^k$ coefficient of $P_1$ and $P_2$ are both zero means
  \[ r_1 b_k + b_{k-1} = r_2 b_k + b_{k-1} = 0 \]
  and hence that $b_k = b_{k-1} = 0$ (since the $r_i$ are nonzero).
\end{proof}

To solve the problem, we use:
\begin{lemma*}
  If $F(x) \in \RR[x]$ is a polynomial with two consecutive zero coefficients,
  it cannot have all distinct real roots.
\end{lemma*}
Here are two possible proofs of the lemma I know (there are more).

\begin{proof}
  [First proof using Rolle's theorem]
  Say $x^t$ and $x^{t+1}$ coefficients of $F$ are both zero.

  Assume for contradiction all the roots of $F$ are real and distinct.
  Then by Rolle's theorem, every higher-order derivative of $F$
  should have this property too.
  However, the $t$th order derivative of $F$ has a double root of $0$, contradiction.
\end{proof}

\begin{proof}
  [Second proof using Descartes rule of signs]
  The number of (nonzero) roots of $F$ is bounded above by
  the number of sign changes of $F(x)$ (for the positive roots)
  and the number of sign changes of $F(-x)$ (for the negative roots).
  Now consider each pair of consecutive nonzero coefficients in $F$,
  say ${\star} x^i$ and ${\star} x^j$ for $i > j$.
  \begin{itemize}
    \ii If $i-j=1$, then this sign change will only count for one of $F(x)$ or $F(-x)$
    \ii If $i-j \ge 2$, then the sign change could count towards both $F(x)$ or $F(-x)$
    (i.e.\ counted twice), but also there is at least one zero coefficient between them.
  \end{itemize}
  Hence if $b$ is the number of nonzero coefficients of $F$,
  and $z$ is the number of \emph{consecutive runs} of zero coefficients of $F$,
  then the number of real roots is bounded above by
  \[ 1 \cdot (b-1-z) + 2 \cdot z = b - 1 + z \le \deg F. \]
  However, if $F$ has \emph{two} consecutive zero coefficients, then the inequality is strict.
\end{proof}

\begin{remark*}
  The final claim has appeared before apparently
  in the HUST Team Selection Test for the Vietnamese Math Society's
  undergraduate olympiad; see \url{https://aops.com/community/p33893374} for citation.
\end{remark*}