Monic polynomial of degree 5

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?

Solution 1:

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) $$f(x)=x(x+4)^2(x-4)^2$$ $$f(1)=225$$

Solution 2:

Since the function attains only positive values for $x > 0$, the root $4$ must be a double root. Since the function is symmetric with respect to the $y$-axis, it also must have $-4$ as a double root. The fifth root is therefore invariant under $x \mapsto -x$, hence is $0$. This implies $f(1) = 225$.