# Factoring a cube

1. Sep 25, 2008

### hayesk85

1. The problem statement, all variables and given/known data

I am trying to find out when the denominator of this equation is zero so I can tell when the graph has asymptotes or holes. For squares I factor such as x2+2x-15 = (x+5) (x-3). How do I do that with a cube?

2. Relevant equations

(x-5) (x+3)
X3-5x2+x-5

2. Sep 25, 2008

### JG89

Have you tried the factor theorem?

Take factors of the last coefficient, which in this case is -5. So you have +/- 1,5 as factors. Substitute those into the polynomial to see which give you 0. In this case it turns out to be 5, s (x-5) is a factor. Now divide X^3-5x^2+x-5 by (x-5), which will give you a quadratic with a remainder of 0. You may or may not be able to further factor the quadratic.

3. Sep 25, 2008

### Dick

Sometimes you can factor just by looking at it, I see x^2*(x-5)+(x-5) right away. In more complicated cases it's handy to use the fact if you write the f(x)=the polynomial, then f(a)=0 means (x-a) is a factor. Divide it out and then try to factor what's left. For easily guessing what might be a root, look up the 'rational roots theorem'.

4. Sep 25, 2008

Thank you!