Ihave no idea what this is! my teacher handed it to me and left.. me !
n! = n(n-1)(n-2)…1

1 Simplify
n
(
n
−
1
)
(
n
−
2
)
×
1
n(n−1)(n−2)×1 to
n
(
n
−
1
)
(
n
−
2
)
n(n−1)(n−2)
n
=
n
(
n
−
1
)
(
n
−
2
)
n=n(n−1)(n−2)

2 Expand
n
=
n
3
−
2
n
2
−
n
2
+
2
n
n=n
3
−2n
2
−n
2
+2n

3 Simplify
n
3
−
2
n
2
−
n
2
+
2
n
n
3
−2n
2
−n
2
+2n to
n
3
−
3
n
2
+
2
n
n
3
−3n
2
+2n
n
=
n
3
−
3
n
2
+
2
n
n=n
3
−3n
2
+2n

4 Move all terms to one side
n
−
n
3
+
3
n
2
−
2
n
=
0
n−n
3
+3n
2
−2n=0

5 Simplify
n
−
n
3
+
3
n
2
−
2
n
n−n
3
+3n
2
−2n to
−
n
−
n
3
+
3
n
2
−n−n
3
+3n
2

−
n
−
n
3
+
3
n
2
=
0
−n−n
3
+3n
2
=0

6 Factor out the common term
n
n
−
n
(
1
+
n
2
−
3
n
)
=
0
−n(1+n
2
−3n)=0

7 Multiply both sides by
−
1
−1
n
(
1
+
n
2
−
3
n
)
=
0
n(1+n
2
−3n)=0

8 Solve for
n
n
n
=
0
n=0

9 Apply the Quadratic FormulaHow?
n
=
3
+
5
2
,
3
−
5
2
n=
2

3+√
5


,
2

3−√
5




10 Collect all solutions from the previous steps
n
=
0
,
3
+
5
2
,
3
−
5
2
n=0,
2

3+√
5


,
2

3−√
5




Done

Ithink the answer is: it is likely.