Solving for $x$ in $\frac{(x+2)!}{(x-1)!}=60$

Question

So I was yet again browsing the homepage of Youtube when I found this video by infyGyan which proposed the following question that I thought that I might be able to solve. The question was $$\text{Solve for }x\text{: }\frac{(x+2)!}{(x-1)!}=60$$Here is my attempt at solving the equation:$\color{white}{\require{cancel}{.}}$ $$\frac{(x+2)!}{(x-1)!}=60$$$$\frac{(x+2)(x+1)(x)(x-1)\dots}{(x-1)(x-2)(x-3)\dots}=60$$$$\frac{(x+2)(x+1)(x)\cancel{(x-1)\dots}}{\cancel{(x-1)(x-2)(x-3)\dots}}=60$$$$(x+2)(x+1)(x)=60$$$$\text{Which simplifies to }x^3+3x^2+2x=60$$Which now we can just plug in numbers to find where both sides equal $60$:$$\text{Test: }1$$$$(1)^3+3(1)^2+2=60$$$$6\neq60$$$$\text{Test: }2$$$$8+12+4=24\neq60$$$$\text{Test: }3$$$$27(2)+6=54+6=60$$$$\therefore x=3$$And plugging it into the original equation gets us$$\frac{5!}{2}=\frac{120}{2}=60$$

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$$\mathbf{\text{My question}}$$

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Is my solution correct, or is there anything I could do to attain the correct solution or attain it more easily?

Answer 1

$$(x+2)(x+1)(x)=60$$

Note $x, x+1, x+2$ are consecutive (adjacent) integers, and $60=2\cdot 2\cdot 3\cdot 5=3\cdot 4\cdot 5$, which is the unique way to factorize $60$ as the product of three consecutive integers.

Therefore, $x=3$

Answer 2

Another way to solve it based on the OP's approach.

By the rational root theorem, we find out that $x = 3$ is a root of the obtained equation.

Taking advantage of such fact, one concludes that:

\begin{align*} x^{3} + 3x^{2} + 2x = 60 & \Longleftrightarrow x^{3} + 3x^{2} + 2x - 60 = 0\\\\ & \Longleftrightarrow (x^{3} - 3x^{2}) + (6x^{2} - 18x) + (20x - 60) = 0\\\\ & \Longleftrightarrow x^{2}(x - 3) + 6x(x - 3) + 20(x -3) = 0\\\\ & \Longleftrightarrow (x^{2} + 6x + 20)(x - 3) = 0\\\\ & \Longleftrightarrow ((x + 3)^{2} + 11)(x - 3) = 0\\\\ & \Longleftrightarrow x = 3. \end{align*}

In other words, the only solution to proposed exercise is indeed $x = 3$.