# Factoring something with cube power?

1. Feb 1, 2010

### zeion

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

I need to find where these two intersect:
x = y^3
x = 3y + 2

2. Relevant equations

3. The attempt at a solution

I set them to equate:
y^3 = 3y + 2
y^3 - 3y + 2 = 0

How do I factor this??

2. Feb 1, 2010

### zeion

I just realized I can sub in 2 to get a factor..
But is there a faster way than to guess randomly?

3. Feb 1, 2010

### dacruick

you cant factor that. But what you could do is try to guess the simplest number you can. and if that number were 1, you would be right and see that it is a factor.

For no reason whatsoever you could also try -2, and you would see that it is also a factor

but thats just a quick evaluation, if you wanted the third factor you could use division of polynomials and hopefully you would get a 2nd degree polynomial which is easily factored

4. Feb 1, 2010

### dacruick

2 is not a factor

5. Feb 1, 2010

### willem2

I suppose he meant: if you substitute y=2 in the polynomial you get 0, so (y-2) has to be a factor.

You can use the rational root theorem to find all rational numbers that might be a factor, in this case you only have to try -2,-1,1 and 2

6. Feb 1, 2010

### zeion

Ah sorry I made a mistake in the equation.. it's supposed to be
y^3 = 3y + 2
y^3 - 3y - 2 = 0

7. Feb 1, 2010

### zeion

What is rational root theorem?
How did you know those numbers to try?
I can always just do a long division right?

8. Feb 1, 2010

### Staff: Mentor

There's a theorem about rational roots of polynomials. For a polynomial anxn + ... + a1x + a0 = 0, if x = p/q is a zero, then p must divide a0 and q must divide an.

With your polynomial, y3 - 3y + 2 = 0, where p/q is a rational root, p has to divide 2 and q has to divide 1. The possible choices for p are +/-1 and +/-2. This means that the possible rational zeroes are +/- 1 and +/- 2. No other rational zeroes are possible. I found that 1 is a root, meaning that y - 1 is a factor.

9. Feb 1, 2010

### zeion

So if I have a missing term in a polynomial I just have to try to factor with the +/- of the first and last coefficients?

10. Feb 1, 2010

### dacruick

y-2 is not a factor still. y+2 is a factor however.