Factoring 40x^3 - 5

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Luke53
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Factoring 40x^3 - 5

Postby Luke53 » Sun Mar 13, 2011 10:00 am

Hello, I'am having problems with factoring 40X^3 - 5 can anyone show me how to do this?

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Martingale
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Re: Factoring

Postby Martingale » Sun Mar 13, 2011 4:08 pm

Luke53 wrote:Hello, I'am having problems with factoring 40X^3 - 5 can anyone show me how to do this?



http://en.wikipedia.org/wiki/Factorization#Sum.2Fdifference_of_two_cubes

Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
Contact:

Re: Factoring

Postby Luke53 » Sun Mar 13, 2011 8:23 pm

OK, thanks for the relply, I know thanks to the Wikipedia factoring page you send me, how to solve this problem now.
I divide each term by 5 to get 8x^3 - 1 that turns in to (2x)^3 - 1^3, then 2x - 1 is a factor, the other factor is found by long division of (2x^3 - 1)/ (2x - 1) = 4x^2 + 2x + 1. I now have to multiply one of the factors by 5 again to make up for the first division by 5. So 40x^3 - 5 = 5 (2x -1) (4x^2 + 2x + 1).
Thank you very much for your help.
Greetings.
Luke.


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