Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
 = 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
 = 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
 A: When you take the $90$ to the other side you need to subtract it. You'll get
$n^2 +7n +12 -90 = 0,$
$n^2 +7n -78 = 0,$
$(n - 6)(n+13) = 0,$
So $n = 6$ or $n = -13$.
Clearly $n = -13 $  doesn't make sense, so we conclude:
$n = 6$.
A: Well, you know that $$\frac{(n+4)!}{(n+2)!} = (n+4)\cdot(n+3) = n^2+7n+12=90$$$$n^2+7n-78=0$$The solutions to this quadratic are $n=6$ and $n=-13$, so we take $n=6$ to be our answer.
A: You can go to the quadratic if you wish, and that solves the problem. But if you simply divide through by $(n+2)!$ you obtain $$(n+3)(n+4)=90$$ and it is easy to see the factorisation of $90$ into two consecutive positive integer factors $(9\times 10)$. The quadratic method does make sure you don't miss the alternative (irrelevant here) $90=-10\times -9$. The fact that it is a quadratic means that you know that there are no more solutions.
A: Just to be different (and to rip off Mark Bennett's answer):
$ab = k> 0$ with $0< a \le b$ means $a \le \sqrt {k}$ and $b \ge \sqrt {k}$. (and if $0< a < b$ then $a < \sqrt k$ and $b > \sqrt{k}$)
So $(n+3)(n+4) = 90$ means $n+3 \le \sqrt{90} \approx 9.smallchange \le n+4$
As $n+3$ and $n+4$ are consecutive we must have:
$(n+3) \le \sqrt{90} < (n+3) + 1$.
Or in other words $n + 3 = \lfloor \sqrt{90} \rfloor = 9$.
So $n = 6$.
