## Solve

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

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

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 Quote by 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
Why the heck did u multiply it??? 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...

## Solve

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.

 Yeah, that's right.