<s>[INST] ARML $1994$ Polynomial Manipulation Question 
(ARML $1994$)
If $x^5 + 5x^4 + 10x^3 + 10x^2 - 5x + 1 = 10$, and $x \neq -1$,
compute the numerical value of $(x + 1)^4$.

I came across this problem in a packet about polynomial manipulation and I couldn't find a way past expanding $(x+1)^4$ and subtracting it from the polynomial. I was wondering if any of you could explain it.
 [/INST] Remember your binomial coefficients, and notice that the left-hand side of the equation is equal to $(x+1)^5-10x$. So we can rewrite that equation as follows:
\begin{align*}
(x+1)^5-10x&=10\\
(x+1)^5-10(x+1)&=0\\
(x+1)\left[(x+1)^4-10\right]&=0 \, .
\end{align*}
Since $x \neq -1$, the first factor above is nonzero, so the second factor must be zero. That is, $(x+1)^4=10$.
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