TITLE: Real valued polynomial has real coefficients
QUESTION [9 upvotes]: If a polynomial has real outputs for all real inputs, it must have all real coefficients. You can see this by using Taylor's formula, and the fact that all derivatives must also have real outputs for real inputs. 
Is there a more immediate proof of this that does not use calculus?

REPLY [2 votes]: Here are two not-so-original approaches that are disguises of some of the ideas in the other answers. Let the polynomial be $p(x)=a_0+a_1x+\cdots+a_{n-1}x^{n-1}$. 

(This one use calculus, though.) Let $D=\operatorname{diag}(1,2,\ldots,n)$. We have
$$
\begin{align*}
&\left(\int_0^1p(x)dx,\int_0^2p(x)dx,\ldots,\int_0^np(x)dx\right)\\
&=\left(a_0,a_1,\ldots,a_n\right)D^{-1}
\pmatrix{
1&1&\cdots&1\\
1&2&\cdots&n\\
1&2^2&\cdots&n^2\\
\vdots&\vdots&&\vdots\\
1&2^{n-1}&\cdots&n^{n-1}
}D.
\end{align*}
$$
Since the vector of integrals is real and both $D$ and the Vandermode matrix are real invertible, each $a_i$ must be real too.
(In the spirit of N.S.'s or lhf's answers.) Pick $n$ distinct points on the real line and construct the Lagrange interpolation polynomial $L$ of $p$ at these $n$ points. As a degree-$(n-1)$ polynomial is uniquely determined by its values at $n$ distinct points, we get $p=L$, which has real coefficients.