How is this possible?

Let $AD=a$ , $BC=b$ , $DF=x-c$ and $FC=c$ . Triangles $\triangle DFE$ and $\triangle DCB$ are similar, hence $\dfrac{x-c}{2}=\dfrac{x}{b}$ . And triangles $\triangle CFE$ and $\triangle CDA$ are also similar, hence $\dfrac{c}{2}=\dfrac{x}{a}$ .

Adding those two equations we get:

$\dfrac{x}{2}=\dfrac{x}{a}+\dfrac{x}{b} \Longrightarrow \dfrac{1}{2}=\dfrac{1}{a}+\dfrac{1}{b}$

Now apply Pythagorean Theorem on triangles $\triangle CDA$ and $\triangle DCB$ :

$a^2+x^2=49 \Longrightarrow a=\sqrt{49-x^2}$

$b^2+x^2=9 \Longrightarrow b=\sqrt{9-x^2}$

Susbtitute $a$ and $b$ on the first equation:

$\dfrac{1}{2}=\dfrac{1}{\sqrt{49-x^2}}+\dfrac{1}{\sqrt{9-x^2}}$

Multiply both sides by $2\sqrt{49-x^2}\sqrt{9-x^2}$ :

$\sqrt{x^4-58x^2+441}=2(\sqrt{49-x^2}+\sqrt{9-x^2})$

Square both sides:

$x^4-58x^2+441=4(58-2x^2+2\sqrt{x^4-58x^2+441})$

$x^4-50x^2+209=8\sqrt{x^4-58x^2+441}$

Square both sides again, simplify, equate to $0$ and find the polynomial:

$x^8+2500x^4+43681-100x^6+418x^4-20900x^2=64x^4-3712x^2+28224$

$x^8-100x^6+2854x^4-17188x^2+15457=0$

$P(x)=x^8-100x^6+2854x^4-17188x^2+15457$

This polynomial is irreducible, hence $P(1)=\boxed{1024}$ .