Solve $\binom{x+2}{5} = 126$ for $x$

To see that there is only one solution, let us generalize slightly and consider the function $f(x)=\binom{x}{k}$ where $x$ is an integer greater than or equal to $k$. We wish to show that this is an increasing function.

In theory, one could express $f(x)$ as a polynomial and use calculus. However, we can take a combinatorial approach.

$f(x+1)$ is the number of ways to select $k$ elements from $\{1,2,\ldots,x+1\}$ We consider two types of subsets, the ones that contain $x+1$ and those that do not. There are $f(x)$ subsets of the second type, and a positive number of the first type (as we are selecting $k-1$ elements from $\{1,2,\ldots,x+1\}$.

Therefore, $f(x+1)>f(x)$ and so $f(x)$ is an increasing function.

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I don't know that one can completely avoid guessing without resorting to calculus, but since the largest prime factor of $126*5!$ is $7$. In particular, $11$, $13$, and $17$, are not factors, so you know that if $x>8$, then $x\geq 20$. Additionally, $x\geq 5$. If you verify that $8$ is too big, that gets you a small range to check.

Hints:

To find an estimate of the root, notice that the RHS is rather large and by $P(x)\approx x^5$, you can expect a root close to $\sqrt[5]{126\cdot5!}=6.85$, so you evaluate at $6$ and $7$.

The derivative of the product is a biquadratic polynomial, so that the extrema can be found and the impossiblity of other real roots confirmed.

Hint:

$$5!*126=9*8*7*6*5$$ $$(x+2)(x+1)(x)(x-1)(x-2)=9*8*7*6*5$$