How to find $P(-1)$ for $\frac{P(2x)}{P(x+1)}=8-\frac{56}{x+7}$ and $P(1)=1$? $P(x)$ is a polynomial such that $P(1)=1$ and $\frac{P(2x)}{P(x+1)}=8-\frac{56}{x+7}$ and $P(-1)$ is rational. How to find $P(-1)$?

$8-\frac{56}{x+7}=\frac{8x+56-56}{x+7}=\frac{8x}{x+7}=\frac{P(2x)}{P(x+1)}$.
Expanding, we get,
$8xP(x+1)=(x+7)P(2x)$
It can be seen that $2x|P(2x)$, so $P(x)=xQ(x)$ for some polynomial $Q$.
Also, $x+7|P(x+1)$, so $P(x)=(x+6)R(x)$ for some polynomial $R$.
Combining the above results, $P(x)=x(x+6)S(x)$. We attempt to substitute it back in.
$\frac{8x}{x+7}=\frac{2x(2x+6)S(2x)}{(x+1)(x+7)S(x+1)}$
Simplifying and multiplying out, we get:
$(2x+2)S(x+1)=(x+3)S(2x)$.
Continuing the argument, we see that $x+2|S(x)$.
As mrprottolo noticed by comparing coefficients it is of degree 3, the polynomial is:
$$P(x)=\alpha x(x+2)(x+6)$$
which can also be verified by substitution. Plugging in $x=1$ to find $\alpha$, we have $\alpha=\frac{1}{21}$. Plugging in $-1$, $P(-1)=\frac{1}{21}\times(-1)\times(2-1)\times(6-1)=-\frac{5}{21}$.
