If $(n + 2)! = 60 [(n - 1)!]$ , find $n$ .

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Solution

$\begin{matrix} (n + 2)! = 60 (n - 1)! \Rightarrow (n + 2) (n + 1) n (n - 1)! = 60 (n - 1)! \Rightarrow n (n + 1) (n + 2) = 60 \Rightarrow n (n^{2} + 3 n + 2) = 60 \Rightarrow n^{3} + 3 n^{2} + 2 n - 60 = 0 \Rightarrow (n - 3) (n^{2} + 6 n + 20) = 0 \end{matrix}$

$n = 3$ , since $n^{2} + 6 n + 20$ has imaginary zeroes
Therefore, $n = 3$

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