Solve

[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

[Coelum]

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

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

[dextercioby]

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

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:

[Mo]

:blushing:

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.

[Nylex]

Yeah, that's right.