TITLE: Monic polynomial of degree 5
QUESTION [12 upvotes]: Consider a monic polynomial $f(x)$ of degree $5$. The graphs of $f(|x|)$ and $|f(x)|$ are same. If $4$ is a root of $f(x)$, find $f(1)$.

I had some problems while drawing the graph of the function. Can anyone explain the solution?

REPLY [21 votes]: If $x\ge0$ then

$f(|x|)=f(x)=|f(x)|$, so $f(x)\ge0$
$f(\left|-x\right|)=f(x)=|f(-x)|$. This, and the fact that $f$ is of odd degree, together imply $f$ is an odd function.

The non-negativity over $\Bbb R^+$, oddness and $f(4)=0$ mean that $+4$ and $-4$ are double roots. Therefore
(since $f$ is monic)
$$f(x)=x(x+4)^2(x-4)^2$$
$$f(1)=225$$