Factorizing a Cubic Equation: How Do I Solve (x-1)(2x-1)(3x-1) = 0?

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To solve the equation (x-1)(2x-1)(3x-1) = 0, one does not need to multiply it out, as the factors can be set to zero directly. The roots can be found by solving each factor: x-1=0 gives x=1, 2x-1=0 gives x=1/2, and 3x-1=0 gives x=1/3. This method is simpler than expanding the equation into a cubic form. The discussion emphasizes the importance of recognizing when an equation is already factored for easier solutions. Overall, the correct values of x are 1, 1/2, and 1/3.
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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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Mo said:
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...
 
Mo said:
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... :-p
 
: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.
 
Yeah, that's right.
 
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