Let Q(x) be a polynomial of degree n. If Q(1) | vedclass

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(C) $Q(x)$ is a polynomial of degree $n$. Given $Q(1)=1$ and $\frac{Q(2x)}{Q(x+1)}+\frac{56}{x+7}-8=0 \dots (i)$. Putting $x=0$ in Eq. $(i)$,we get $\

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$8$

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(C) $Q(x)$ is a polynomial of degree $n$. Given $Q(1)=1$ and $\frac{Q(2x)}{Q(x+1)}+\frac{56}{x+7}-8=0 \dots (i)$. Putting $x=0$ in Eq. $(i)$,we get $\frac{Q(0)}{Q(1)}+\frac{56}{7}-8=0$. Since $Q(1)=1$,we have $Q(0)+8-8=0$,so $Q(0)=0$. Rearranging Eq. $(i)$,we get $\frac{Q(2x)}{Q(x+1)} = 8 - \frac{56}{x+7} = \frac{8x+56-56}{x+7} = \frac{8x}{x+7} \dots (ii)$. Since $Q(0)=0$,$x$ is a factor of $Q(x)$. Let $Q(x) = x P(x)$. Substituting into $(ii)$,$\frac{2x P(2x)}{(x+1) P(x+1)} = \frac{8x}{x+7} \implies \frac{P(2x)}{P(x+1)} = \frac{4(x+1)}{x+7}$. For $x=-1$,$P(-2)=0$,so $(x+2)$ is a factor of $P(x)$. Let $P(x) = (x+2) R(x)$. Then $\frac{(2x+2) R(2x)}{(x+3) R(x+1)} = \frac{4(x+1)}{x+7} \implies \frac{R(2x)}{R(x+1)} = \frac{2(x+3)}{x+7}$. For $x=-3$,$R(-6)=0$,so $(x+6)$ is a factor of $R(x)$. Let $R(x) = (x+6) S(x)$. Then $\frac{(2x+6) S(2x)}{(x+7) S(x+1)} = \frac{2(x+3)}{x+7} \implies \frac{S(2x)}{S(x+1)} = 1$. Thus $S(x)$ is a constant. So $Q(x) = c \cdot x(x+2)(x+6)$. Since $Q(1)=1$,$c(1)(3)(7)=1 \implies c = \frac{1}{21}$. Degree $n=3$. The value of ${}^nC_0+{}^nC_1+\ldots+{}^nC_n = 2^n = 2^3 = 8$.

Let $Q(x)$ be a polynomial of degree $n$. If $Q(1)=1$ and $\frac{Q(2x)}{Q(x+1)}+\frac{56}{x+7}-8=0$,then the value of ${}^nC_0+{}^nC_1+\ldots+{}^nC_n$ is equal to

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