combination equation help Thanks

[marvinfish]

Hi,
I am stuck on this math problem. n c 3 = 10. I know the answer is 5. However I have only been able to simplify the equation to n (n-1) (n-2) = 60. How do I factor this expression to get the answer 5?

Thanks so much for your help

 

[Mrspi]

Hi,
I am stuck on this math problem. n c 3 = 10. I know the answer is 5. However I have only been able to simplify the equation to n (n-1) (n-2) = 60. How do I factor this expression to get the answer 5?

Thanks so much for your help

You could multiply out the expressions on the left side:

n[sup:1f8b78uw]3[/sup:1f8b78uw] - 3n[sup:1f8b78uw]2[/sup:1f8b78uw] + 2n = 60

Subtract 60 from both sides:

n[sup:1f8b78uw]3[/sup:1f8b78uw] - 3n[sup:1f8b78uw]2[/sup:1f8b78uw] + 2n - 60 = 0

Now, you have a BIG clue that n = 5 is a solution....(be sure to check to see that it is!!)

Without that clue, you could use the rational roots theorem...any rational solution for this equation must be among these numbers:

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

Since the value of the polynomial is 0 when n = 5:

5[sup:1f8b78uw]3[/sup:1f8b78uw] - 3(5)[sup:1f8b78uw]2[/sup:1f8b78uw] + 2(5) - 60 = 125 - 3*25 + 10 - 60 = 125 - 75 + 10 - 60 = 0, you know that 5 is a solution.

You can use either polynomial long division or synthetic division to show that there are no other REAL solutions.