# How to solve a factorization

1. Dec 18, 2004

### Mo

I know how to solve quadratics using both factorisation and the equation method ... but how can i solve :

$$(x-1)(2x-1)(3x-1) = 0$$

I multiplied it all out and i got ..

$$6x^3 - 2x^2 -3x -1=0$$

I just do not know where to got from here .. a little nudge in the right direction would be appreciated!

regards,
Mo

2. Dec 18, 2004

### Coelum

A small hint: why do you multiply? After all, if a*b*c=0, then....

3. Dec 18, 2004

### dextercioby

Why the heck did u multiply it??? :surprised It was already solved.U were being asked for the 3 possible values of "x" which cancel the expression from the LHS.I think/hope they were obvious...

Daniel.

PS.But if u'd rather apply Cardano's formulae for the 3rd order algebrac equation u got,be my guest... :tongue2:

4. Dec 18, 2004

### Mo

yes, i do understand.

if

(x-1) = 0 then x =1
(2x-1) =0 then 2x=1 so x = 1/2
(3x-1) = 0 then 3x=1 so x=1/3

is that right.Am i going about it right. thanks.

5. Dec 18, 2004

### Nylex

Yeah, that's right.