factoring 18r^3t^2 + 12r^2t^2 - 6r^2t  TOPIC_SOLVED

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factoring 18r^3t^2 + 12r^2t^2 - 6r^2t

Postby Motherof8 on Tue Jan 17, 2012 9:32 pm

I am trying to factor 18r cubed t squared + 12 r squared t squared - 6 r squared t. When I try (3 r squared t + 2t) (6rt -3r), it works out except I get -9r squared t in the middle. Is there any other way to factor this? I'd like an answer as soon as possible. Thank you.
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Postby stapel_eliz on Wed Jan 18, 2012 3:27 am

I'm not sure how you're doing your factoring...?

The first step should be taking the common factor out front. With what does this leave you? Is that particularly factorable? :wink:
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Re: factoring 18r^3t^2 + 12r^2t^2 - 6r^2t

Postby Motherof8 on Wed Jan 18, 2012 5:02 am

I tried factoring it again and everything checks out except I get -9r cubed t (-9r^3t) in the middle. I want to know how factor the equation so as to get rid of the 9, since it's obviously not in the final answer. Is it (3r^2t +2rt) (6rt -3r) or something else?
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Postby stapel_eliz on Wed Jan 18, 2012 12:58 pm

Motherof8 wrote:I tried factoring it again....

What is the reasoning and methodology of your factoring? What are you doing when you're factoring? What are your steps?

Motherof8 wrote:I want to know how factor the [expression]

Please take another look at the previous reply. :wink:
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Re: factoring 18r^3t^2 + 12r^2t^2 - 6r^2t  TOPIC_SOLVED

Postby Motherof8 on Wed Jan 18, 2012 3:36 pm

Eureka! (I have read that eureka means "I have found it.") I believe I found the factors. Instead of 3 and 6, they are 2 and 9. I surmise that there was a typo in the book and -6rt^2 was meant. If one puts 2 r cubed t squared + 2rt x 9rt -3t,the answer works out. I don't need any more help for this problem.
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