10. Suppose $P(x)$ is a polynomial such that $P(1)=1$ and
$$
\frac{P(2 x)}{P(x+1)}=8-\frac{56}{x+7}
$$
for all real $x$ for which both sides are defined. Find $P(-1)$.
Solution:
$-5 / 21$

Cross-multiplying gives $(x+7) P(2 x)=8 x P(x+1)$. If $P$ has degree $n$ and leading coefficient $c$, then the leading coefficients of the two sides are $2^{n} c$ and $8 c$, so $n=3$. Now $x=0$ is a root of the right-hand side, so it's a root of the left-hand side, so that $P(x)=x Q(x)$ for some polynomial $Q \Rightarrow 2 x(x+7) Q(2 x)=8 x(x+1) Q(x+1)$ or $(x+7) Q(2 x)=4(x+1) Q(x+1)$. Similarly, we see that $x=-1$ is a root of the left-hand side, giving $Q(x)=(x+2) R(x)$ for some polynomial $R \Rightarrow 2(x+1)(x+7) R(2 x)=$ $4(x+1)(x+3) R(x+1)$, or $(x+7) R(2 x)=2(x+3) R(x+1)$. Now $x=-3$ is a root of the left-hand side, so $R(x)=(x+6) S(x)$ for some polynomial $S$.
At this point, $P(x)=x(x+2)(x+6) S(x)$, but $P$ has degree 3 , so $S$ must be a constant. Since $P(1)=1$, we get $S=1 / 21$, and then $P(-1)=(-1)(1)(5) / 21=-5 / 21$.