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 $P(-1)$?
 A: $8-\frac{56}{x+7}=\frac{8x+56-56}{x+7}=\frac{8x}{x+7}=\frac{P(2x)}{P(x+1)}$.
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 $-1$, $P(-1)=\frac{1}{21}\times(-1)\times(2-1)\times(6-1)=-\frac{5}{21}$.
A: The other answer already answers the question, but just because it was not obvious to me, I want to explain how you get that the polynomial is of degree $3$. 
Assume that the degree of $P(x)$ is $k$, i.e. $$P(x)=a_kx^k+a_{k-1}x^{k-1}+\dots+a_1x$$ with $a_k\neq0$. There is no constant term, because one can easily see that $P(0)=0$ (substitute $x=0$ in the original equation). So, starting from $$8xP(x+1)=(x+7)P(2x)$$ consider only the coefficient of the highest term, i.e. $x^k$. On the LHS we have $$8xP(x+1)=8x(a_kx^k+\ldots)=8a_kx^{k+1}+\ldots$$ and on the RHS we have $$(x+7)P(2x)=(x+7)(a_k(2x)^k+\ldots)=a_k2^kx^{k+1}+\ldots$$ So, now comparing coefficients yields $$8a_k=2^ka_k\overset{a_k\neq 0}\implies 8=2^k\implies k=3$$ So, we found that $P(x)$ is of degree $3$ and has no constant term (because $P(0)=0$), that is $$P(x)=a_3x(x-c_1)(x-c_2)$$ Now we can use that $P(0)=0$ and $P(1)=1$ in order to obtain the other two roots of the polynomial from the original equation. Choose $x=-1$ to obtain that $$8(-1)P(-1+1)=(-1+7)P(2(-1)) \implies -8P(0)=6P(-2) \implies P(-2)=0$$ and then choose $x=-3$ to obtain that $$8(-3)P(-3+1)=(-3+7)P(2(-3))\implies -24P(-2)=4P(-6)\implies P(-6)=0$$ So, we found that $$P(x)=a_3x(x+2)(x+6)$$ and from $P(1)=1$ we can found that $a_3=1/21$ and therefore that $P(-1)=-5/21$.

Indeed I do not see, where the assumption $P(-1)$ rational, was used.
