Question 200686
The number of ways one can select three cards from a group of n cards (assuming the order of the selection matters), where {{{n>3}}} {{{P(n)=n^3-3n^2+2n}}} , is given by . For a certain card trick a magician has determined that there are exactly 60 ways to choose three cards from a given group. How many cards are in the group?
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Substitute {{{60}}} for {{{P(n)}}} and solve for {{{n}}}:

{{{P(n)=n^3-3n^2+2n}}}
{{{60=n^3-3n^2+2n}}}

Switch sides of the equation:

{{{n^3-3n^2+2n=60}}}

Get 0 on the right by subtracting 60 from both sides:

{{{n^3-3n^2+2n-60=0}}}

The possible rational zeros of P(n) are ± the divisors of 
60.

±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±30, ±60

Try 1 using synthetic division:

1 | 1 -3  2 -60
  |    1 -2   0  
   ------------
    1 -2  0 -60

Nope

Try 2 using synthetic division:

2 | 1 -3  2 -60
  |    2 -2   0  
   ------------
    1 -1  0 -60

Nope

Try 3 using synthetic division:

3 | 1 -3  2 -60
  |    3  0   6  
   ------------
    1  0  2 -54

Nope

Try 4 using synthetic division:

4 | 1 -3  2 -60
  |    4  4  24  
   ------------
    1  1  6 -36

Nope

Try 5 using synthetic division:

5 | 1 -3  2 -60
  |    5 10  60  
   ------------
    1  2 12   0

That leaves a 0 remainder, so n=5 is a solution

So we have factored the polynomial equation

 {{{n^3-3n^2+2n-60=0}}}  as

 {{{(n-5)(n^2+2n+12)=0}}}

This gives

{{{n-5=0}}} and {{{n^2+2n+12=0}}}

{{{n=5}}}

There can be no positive solutions of
{{{n^2+2n+12=0}}} since there are
no sign changes.

So the answer is n=5.

Edwin</pre>