# factor 3x^3 - 125



## daaaaave (Jan 1, 2007)

what is the factorization of this? :con 

3x^3 - 125

one of the reasons i got fired was for not knowing how to do this ops

i feel like it can't really be factored

also how would you factorize 3x^2 - 5x + 10

I got (3x - 2) (x - 1) + 8

Is that the simplest form? I also could not explain that well as to how I used completing the square method


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## Bon (Dec 24, 2005)

What type of work were you doing?


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## daaaaave (Jan 1, 2007)

i am a tutor. i lost a client


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## pariahgirl (Mar 26, 2008)

ah sorry i didn't see the three in front..i tried factoring this but i don't honestly see a way to do it because x comes out to be an irrational number..


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## daaaaave (Jan 1, 2007)

How'd you do the first step to get (x-5)?

Also, you're expression expands to x^3 - 125 not 3x^3 - 125


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## Ignivomous (Mar 31, 2008)

As far as I know, that cannot be factored. The formula for factoring a difference of cubes is: a^3 - b^3 = (a - b)(a^2 + ab + b^2). The coefficient of 3 kills that. I'm not really sure why you'd want to factor it anyway; it's already simplified. If you turn the expression into an equation by setting it equal to zero, you end up with x = 5/cubed root(3). No factoring needed.

For the second, again, I would turn the expression into an equation with the intent of solving for x, which is why you'd want to factor. Setting equal to zero and completing the square gave me [x - (5/6)]^2 + (95/36) = 0, which is actually a complex number when solved, so you'd have to set the original expression equal to at least (95/12) in order to get a real answer (x = 5/6 in that case).


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## daaaaave (Jan 1, 2007)

Thanks, guys. The more I think about this, the guy should have given me some context as to the kind of work we'd be doing. He could not factor x^2 - 36 so I see no way he is going to be able to get through the course he's taking especially if its only the first month and he's doing problems with imaginary roots and cubic factoring.


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## Zephyr (Nov 8, 2003)

Are you sure the numbers were right? The first would be a nice difference of cubes if only it were x^3 - 125. If it were 3x^3 - 375, then you could pull out the 3 to get 3(x^3 - 125) and then factor the stuff in the brackets. As it is it would be really messy to factor it as a difference of cubes. What course it is, exactly?


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## daaaaave (Jan 1, 2007)

I dunno exactly...the book was called Technical Mathematics and it was a prereq for an engineering course he said covering algebra, trig, and calc.

I wrote the problems down from this worksheet so that's what they were, I dunno.


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