Question

What is the greatest common factor of 6x^4 - 12x^3?

Answer 1

The greatest common factor is the largest number and largest variable you can "take out" of both terms. Does 6 go into both 6 and 12? If it does, 6 is your greatest common factor. What about the x's? What is the greatest number of x's you can factor out of both terms?

Answer 2

Yes, 6 does go into both 6 and 12. So that is the greatest common factor? I have no clue how to do the x's though..?

Answer 3

The greatest number of x's that can be factored out of both terms is either 3 or 2. Or is that wrong?

Answer 4

Like this: when you factor out a 6 from the first term, you are left with 1; when you factor a 6 out of 12 in the second term you are left with 2. Now let's do the x's:

Answer 5

\[x ^{4}=x ^{1}*x ^{1}*x ^{1}*x ^{1}\]

Answer 6

\[x ^{3}=x ^{1}*x ^{1}*x ^{1}\]

Answer 7

In order to factor out the greatest common factor of the x terms, we have to see how many we are "allowed" to take out. The first term is x^4, and the second term is x^3. When you divide x^3 by x^3 you get:

Answer 8

\[\frac{ x ^{1}*x ^{1}*x ^{1} }{ x ^{1}*x ^{1}*x ^{1} }\]Those all cancel out as you can see. When you divide x^3 out of x^4 this is what you get:

Answer 9

\[\frac{ x ^{1}*x ^{1}*x ^{1}*x ^{1} }{ x ^{1}*x ^{1}*x ^{1} }\]You can cancel all three out of the denominator but you are left with one in the numerator. So our greatest common number we can take out is 6, and our greatest common variable we can take out is x^3. Like this:

Answer 10

\[6x ^{4}-12x ^{3}->6x ^{3}(x-2)\]

Answer 11

So \[6x ^{3}\]is your greatest common factor.

Answer 12

Ok, thank you so much, that makes much more sense to me now and helps alot! My next question is to factor the following expression... 3y^3-9y^2+4y-12. Any ideas? Thanks.

Answer 13

you dont get it till now..???@allethaweapon

Answer 14

What do you mean? I understand how to get the greatest common factor of that problem now, but I wanted to read all of those replies and let that person finish explaining before I did anything or said anything else. Now I am on to a new question, as I asked right before you replied.

Answer 15

Good job, @IMStuck and @zaibali.qasmi