Finding GCF of 3 Terms: 102k^5m^2, 51k^4m, 153k^2m^2

  • #1
dizco29
32
0
hey guys, getting back into some basic algebra (so hope you don't mind a real newbie) and was doing a factoring tutorial on the web, and came across this.

The work for finding the GCF of three terms is shown below.

102k^5m^2
51k^4m
153k^2m^2


First find the GCF of the coefficients:
102 (1, 2, 3, 6, 17, 34, 51, 102)
51 (1, 3, 17, 51)
153 (1, 3, 9, 17, 51, 153)
GCF (of coefficients only) = 51


Next find the GCF of the variables:

k^5 m^2

k^4 m

k^2 m^2

GCF (of variables) = k2m


Now multiply the two GCFs
GCF of the entire term = 51k^2 m

what I don't understand is how the GCF for K is 2. Because 5 is not divisble by 2. So how can that be?

here's a link to the page if the above is not clear.

http://www.algebrahelp.com/lessons/f...indgcf/pg2.htm

Thanks!
 
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  • #2
k^5 = k x k x k x k x k = k^2 x k^3. Does that help?
 
  • #3
hmmm, kinda, can you explain a little bit more about it. didn't know that you can add the exponents when you are looking for the GCF
 
  • #4
Let us denote k^2 by n. Therefore k^5 = n.k^3 and k^2 = n. Therefore n is the GCF.
 
  • #5
  • #6
cool, I get it now! stupid me lol. thanks for your explainantion!
 
  • #7
dizco29 said:
I still don't get how in this case the common factor for z is ^2. shouldn't it be z^1?
thanks!
z^3 = z^2 x z
 
  • #8
dizco29 said:
cool, I get it now! stupid me lol. thanks for your explainantion!
You're welcome. :smile:
 
  • #9
dizco29 said:
what I don't understand is how the GCF for K is 2. Because 5 is not divisble by 2. So how can that be?

You're error is clear in the wording- "The GCF for K" is not 2! It is k2. 5 is not divisible by 2 but k5 certainly is divisible by k2.
 
  • #10
this is a slly problem to do by that method. this poroblem involves only the same prime factors. e.g. to find the gcf of expressions of form a^n b^m c^p, where a,b,c, are all prime, just take each prime and raise it to the smallest power it has in any term given.

e,g, in your example you have 3^r 17^s k^t m^u, and just taking the smallest power of each occurrence gives immediately 3^1 17^1 k^2 m^1.
 

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