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Factorization #2

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Factorization #2

by C » Thu Feb 25, 2010 2:07 am
I'm looking for an effective way to factor the following algebra problems:

x^3-3x^2+2x

or

n^2-14n+49

I can factor these given some intuitive thinking and try-fail-method, but it takes me too much time.

Does anyone have a good method for this kind of factoring?

Thanks!
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by Pedros » Thu Feb 25, 2010 2:15 am
X^3 - 3x^2 + 2x

x is common in the three terms so lets factor it out
x ( x^2 - 3x +2 )

the term inside the brakets could be factored as well

x ( x-2 ) ( x-1)

you have a product of 3 consecutive integers here if X is an integer
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by Pedros » Thu Feb 25, 2010 2:20 am
n^2 - 14n + 49

only 7*7 gives 49

so its ( n-7)(n-7) or ( n-7)^2

Trial and error is ok , after some pratice you will be able to factor them pretty quickly.
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