# Homework Help: Complex roots

1. Oct 22, 2007

### Corkery

1. The problem statement, all variables and given/known data
6x^3 - 3x^2 - 45x

2. Relevant equations

3. The attempt at a solution
-first factor out 3x
3x(2x^2 - x -15)

2 x 5 = 30 so...

3x(2x^2 + 5x - 6x - 15)

-separate the equations
3x(2x^2 + 5x)( - 6x -15)

-simplify a few things.
3x^2(2x + 5) -3 (2x + 5)

(3x^2 - 3)(2x + 5)

set both equations to zero
2x + 5 = 0
2x = -5
x = -5/2

3x^2 - 3 = 0
3x^2 = 3

thats where I get stuck, that is if I did this right. thanks for any help you can offer.

2. Oct 22, 2007

### cristo

Staff Emeritus
What've you done to get from the line above to this:
It's incorrect, anyway. You should try using the quadratic formula to factorise (2x^2 -x- 15)

3. Oct 23, 2007

### HallsofIvy

Wasn't this supposed to be an equation? And, if so, doesn't it need an "=" somewhere? Did you mean 6x^3- 3x^2- 45x= 0?

Well, no, 2 x 5= 10, not 30. But there is no 30 in the equation anyway so I don't know what you were trying to do!

?? Did you mean 3x[(2x^2+ 5x)+ (-6x-15)]?

Again, you mean 3x^2[(2x+5)- 3(2x+5)]

No, since the 3x is multiplied by both 2x+ 5 and -3(2x+5) you cannot separate the expression like that.
Well, obviously that gives x^2= 1 which has solutions x= 1 and x= -1- but they obviously do not satisfy the original equation. Go back to 3x^2(2x^2 - x -15)= 0. That gives you one obvious solution. Now try to factor 2x^2- x- 15.

However, this equation does not has complex roots!

Last edited by a moderator: Oct 23, 2007
4. Oct 24, 2007

### Corkery

ok thanks a lot. that really does help a lot. My tutor also pointed out some of my mistakes, but with the combination of that and this I'm very clear on the subject. thanks again