An algebra problem by Cameron CHang

Let

$\begin{aligned} P & = \frac {4^4+4^3-7\cdot 4^2-4+6}{4^4-13\cdot 4^2+36}\cdot \frac {5^4+5^3-7\cdot 5^2-5+6}{5^4-13\cdot 5^2+36} \cdots \frac {n^4+n^3-7n^2-n+6}{n^4-13n^2+36} \\ & = \prod_{k=4}^n \frac {k^4+k^3-7k^2-k+6}{k^4-13 k^2+36} \\ & = \prod_{k=4}^n \frac {(k-1)(k+1)(k-2)(k+3)}{(k^2-4)(k^2-9)} \\ & = \prod_{k=4}^n \frac {(k-1)(k+1)(k-2)(k+3)}{(k-2)(k+2)(k-3)(k+3)} \\ & = \prod_{k=4}^n \frac {(k-1)(k+1) }{(k-3)(k+2)} \\ & = \prod_{\color{#3D99F6}k=1}^{\color{#3D99F6}n-3} \frac {(k+2)(k+4)}{k(k+5)} \quad \quad \small \color{#3D99F6} \text{Putting }n=117 \\ & = \prod_{\color{#3D99F6}k=1}^{\color{#3D99F6}114} \frac {(k+2)(k+4)}{k(k+5)} \\ & = \frac {5!116!118!}{2!4!114!119!} \\ & = \frac {5\cdot 115 \cdot 116}{2\cdot 119} \end{aligned}$

$\implies \left(4\cdot 4! + 3\cdot 3! + 2\cdot 2! + 1\cdot 1! \right) P = 119 \cdot \dfrac {5\cdot 115 \cdot 116}{2\cdot 119} = \boxed{33350}$