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Mathematics 21 Online
OpenStudy (anonymous):

Find the GCF. A. 28a3b3c4 The three's and the four are small here. B. 14 C. 14abc D. 28abc

OpenStudy (anonymous):

OpenStudy (anonymous):

The equation is above.

OpenStudy (phi):

can you find the GCF for 14 and 28?

OpenStudy (phi):

GCF is the biggest number that divides evenly into both.

OpenStudy (anonymous):

7? @phi

OpenStudy (phi):

7 is a common factor, but not the biggest. how about 14?

OpenStudy (phi):

14 divides into 14 and into 28 evenly, right?

OpenStudy (anonymous):

That would work!

OpenStudy (phi):

so 14 is part of the answer. any idea what the GCF is for a and a*a ?

OpenStudy (anonymous):

That's where I get lost! lol

OpenStudy (phi):

with a letter you don't have a lot of options. the only thing we know that will divide evenly into "a" is itself.

OpenStudy (anonymous):

and what is A. standing for?

OpenStudy (phi):

a is a number. But we know this: a/a = 1 (anything divided by itself is 1) which means a divides evenly into a. The next question is , does a divide evenly into a*a ?

OpenStudy (anonymous):

Yeah I think it would.

OpenStudy (phi):

you can re-write it like this \[ \frac{a\cdot a}{a} = \frac{a}{a} \cdot a\] now use a/a= 1 to say \[ 1 \cdot a = a \] in other words, a*a/a = a with NO remainder. (i.e. divides evenly) so a divides evenly into a and into a*a and because there is no other thing (except 1) that divides into a (as far as we know), a is the greatest common factor of a and a*a

OpenStudy (phi):

btw, \[ a \cdot a= a^2\] so we found the GCF of \( a \text{ and } a^2\)

OpenStudy (phi):

so far the GCF is 14a any idea what the GCF of b and b*b is ?

OpenStudy (anonymous):

7?

OpenStudy (phi):

the gcf of a letter b and b*b is not a fixed number. It will be a letter

OpenStudy (anonymous):

b?

OpenStudy (anonymous):

LOL Im sorry... I'm horible at these thing's.

OpenStudy (phi):

yes, b is the only thing other than 1 that evenly divides into both b and b*b

OpenStudy (anonymous):

YAY!

OpenStudy (phi):

If you are lost, it's because there is an idea you are missing. Once you get the idea, it makes sense.

OpenStudy (phi):

now the gcf of c and c*c*c ?

OpenStudy (anonymous):

C.!

OpenStudy (anonymous):

Option C. is looking promising...

OpenStudy (phi):

yes, and all together we get 14abc

OpenStudy (anonymous):

WOW!! Thanks for working with me man! I really appreciate it!!

OpenStudy (anonymous):

*handshake*

OpenStudy (phi):

if you have the idea, then what is the gcf of a*a and a*a*a ? any idea ?

OpenStudy (anonymous):

A.!

OpenStudy (phi):

a is a factor, but we can do more.

OpenStudy (anonymous):

A and A???

OpenStudy (phi):

\[ \frac{a\cdot a}{a\cdot a} = \frac{a}{a} \cdot \frac{a}{a} = 1\] in other words a*a divides evenly into a*a

OpenStudy (phi):

yes, but I would say A TIMES A (not A and A)

OpenStudy (anonymous):

OOOOhhh alright! Cool!

OpenStudy (phi):

and \[ \frac{a\cdot a\cdot a}{a\cdot a} = a \] so a*a is the gcf of a*a and a*a*a

OpenStudy (phi):

or, using exponents the gcf of \( a^2 \) and \(a^3\) is \( a^2 \)

OpenStudy (anonymous):

Okay! Wow. Thanks again for all your brainpower! =)

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