Find the value of $Q(x)$ at $x= -1$, knowing some of its properties.

Question

Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and $$ \frac{Q(2x)}{Q(x+1)}= 8 -\frac{56}{x+7}, \quad \forall x \ne -7 \text{ and } Q(x+1)\neq 0\,. $$ Find $Q(-1)$.

Answer 1

Rewrite the equation as $$(x+7)\,Q(2x)=8x\,Q(x+1)\,.\tag{*}$$ Thus, $2x\mid Q(2x)$ or $x\mid Q(x)$, and $x+7\mid Q(x+1)$ or $x+6\mid Q(x)$. That is, $Q(x)=x(x+6)\,R(x)$ for some polynomial $R(x)$. Show that $$(x+3)\,R(2x)=2(x+1)\,R(x+1)\,.$$ Ergo, $2x+2\mid R(2x)$ or $x+2\mid R(x)$, and $x+3\mid R(x+1)$ or $x+2\mid R(x)$. Thence, $R(x)=(x+2)\,S(x)$ for some polynomial $S(x)$. Then, prove that $$S(2x)=S(x+1)\,,$$ which implies that $S(x)$ is a constant polynomial. Therefore, $$Q(x)=kx(x+2)(x+6)$$ for some constant $k$.

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You can also notice that $\lim\limits_{x\to\infty}\,\dfrac{Q(2x)}{Q(x+1)}=8=\left(\dfrac{2}{1}\right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0\,,$$ and so on.