Greatest Common Factors

kconstien

New member
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Nov 27, 2006
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I do not understand GCF, do you need to factor down each number for instance
the GCF for 1200 and 450
Would that be 2 or 150?

and for 11b6, 11b4k2, 19b4
would this be b4?

Thanks for your help
 

royhaas

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Dec 14, 2005
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The greatest common factor of 1200 and 450 is 150, since 1200=150*8, 450=150*3, and the greatest common factor of 3 and 8 is 1. You should factor both numbers into a product of prime powers.
 

kconstien

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Nov 27, 2006
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Thank you for that. I wasn't sure if you had to factor all the way down.

Could you help me with the other one.
11b*6, 11b*4K*2, 19b*4

I think this would be b*4 since I don't see anything in commone with the whole number.

I greatly appreciate your help.
Thanks.
 

Denis

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Feb 17, 2004
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kconstien said:
11b*6, 11b*4K*2, 19b*4
I think this would be b*4 since I don't see anything in commone with the whole number.
What do you mean with b*6; b to the power 6? If so, show this way: b^6.
If you mean b times 6, then 11b*6 = 66b: quite a different animal!
 

kconstien

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Nov 27, 2006
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Thank you, I'm a total rookie!

The question is find the GCF of

11b^6, 11b^4k^2, 19b^4

Thanks for your help.
 

stapel

Super Moderator
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Feb 4, 2004
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Follow the same process explained earlier: Factor each term completely. The greatest common factor will be the product of the factors that each term shares.

If you get stuck, please reply showing your work and reasoning, starting with your complete factorization of each term. Thank you.

Eliz.
 

Denis

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Feb 17, 2004
Messages
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kconstien said:
The question is find the GCF of
11b^6, 11b^4k^2, 19b^4
Sometimes easier if you pretend you're adding the terms:
11b^6 + 11b^4k^2 + 19b^4
= b^4(? + ? + ?)
 
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