Question

Divide (18x^3 + 12x – 3x) ÷ 6x^2

Answer 1

Do you know how to factor 18x^3 + 12x – 3x?

Answer 2

not really

Answer 3

Okay, do you know how to find the greatest common factor?

Answer 4

yeah.

Answer 5

So, what's the greatest common factor of 18x^3 + 12x – 3x?

Answer 6

36

Answer 7

wait thats multipule huh?

Answer 8

Haha, yes, I was just going to say that. So do you know how to find the GCF? If not, I can walk you through it.

Answer 9

no

Answer 10

Okay, so first we need to list all the prime factors of the coefficients. The coefficients are 18, 12, and 3.
3: 3
12: 2, 2, 3
18: 2, 3, 3

Now, I want you to tell me what one prime factor all of the coefficients have in common.

Answer 11

Oh yeah I remember this. Okay 3.

Answer 12

That's correct! Now, let's look at our equation again.
18x^3 + 12x – 3x
Since all the numbers have at least one x, we can add x to the GCF.
So the GCF is 3x. Does it make sense how we got that?

Answer 13

Yes

Answer 14

Okay, do you know how to factor the GCF out?

Answer 15

Uhm.. I dont really remember

Answer 16

That's okay! :D
We need to put the GCF on the left side of a set of parenthesis. Like this:
3x( )
And now we need to figure out what to put in those parenthesis. And in order to do that, we need to figure out what we can multiply by 3x to get our original expression, 18x^3 + 12x – 3x. 3x times what equals 18x^3?

Answer 17

6x?

Answer 18

That's almost right. But we need to come up with 18x CUBED, that is 18^3. So in order to do that, we need to multiply 3x by 6x^2 to get 18x^3, correct?

Answer 19

Oh alright. That makes since. So then you put 6x^2 in the parenthesis right?

Answer 20

That's right! So now we have:
[3x(6x^2 )
Remember, our original expression was 18x^3+12x-3x.
We already factored out the 1st term (18x^3), so let's move on to the second term (12x)
So 3x times what equals 12x?

Answer 21

4

Answer 22

That's right!
3x(6^2+4 )
Now, we have factored out the 1st and 2nd term of 18x^3+12x-3x. We just have the 3rd term left (-3x).
3x times what equals -3x?

Answer 23

-1?

Answer 24

Yes, that's correct..
Dang, okay, don't be mad at me, but I think I read your question wrong. At first, we were supposed to combine the like-terms 12x and –3x which would equal 9x. 18x^3+9x.
Are you okay with starting from scratch? It won't take as long this time since you understand the concepts.

Answer 25

Yeah no problem. Your super helpful

Answer 26

Yay, okay! :D
We have 18x^3 + 9x. We need to find the GCF of this expression, just like last time. Do you know how to do that now?

Answer 27

Yeah so it would still be 3 wouldn't it?

Answer 28

Yes, that's a factor of both of them, but not the GREATEST common factor. :D So let's list the prime factors of the coefficients, 18 and 9.
9: 3, 3
18: 2, 3, 3
They both have two 3's in common, right? We would multiply the two 3's to get the number 9. That's the first part of our GCF. Since our terms both have at least one x, we can also add the x to our GCF. So, the GCF is 9x. Does it make sense how we got that?

Answer 29

Yes

Answer 30

Okay, we need to put it on the left side of a set of parenthesis just like lat time:
9x( )
Remember our original expression, 18x^3 + 9x.
9x times what equals 18x^3?

Answer 31

2x^2

Answer 32

Yes!
9x(2x^2 )
Now we factored out our first term (18x^3), and now for our second term (9x).
In order to get 9x from 9x, we just multiply by 1, correct? So, this is our factored form of the expression:
\[9x(2x^2 * 1)\]
That's just the numerator, though. We need to find\[\frac{9x(2x^2+1)}{6x^2}\]

The denominator (6x^2) doesn't need to be factored. Now that we have both the numerator and the denominator factored, we can begin to divide out terms. Do you know how to do that?

Answer 33

Sorry, my first equation is supposed to be 9x(2x^2+1), not times 1. Oops.

Answer 34

You divide like terms right?

Answer 35

That's correct, so can you do that?

Answer 36

So like 2x^2 divided by 6x^2?

Answer 37

\[\frac{9x(2x^2+1)}{6x^2}\]

Do you know what the greatest common factor of 6 and 9 is? In other words, what's the greatest number than can divide into both of them?

Answer 38

3?

Answer 39

Yes! So 9÷3 is 3, and 6÷3 is 2. We can divide those out so we get:\[\frac{3x(2x^2+1)}{2x^2}\]
Do you see that both the numerator and the denominator have at least 1 "x" in them? (The numerator has 1 x, and the denominator has 2 x's.)

Answer 40

That is, at least one x outside of the parenthesis

Answer 41

Yes

Answer 42

Okay, good. So we can cancel out 1 "x" from the numerator and the denominator. \[\frac{3(2x^2+1)}{2x}\]

Since we can't divide anything else out, that would be your answer. Does it make sense how we got that?

Answer 43

Yes. Thanks so much!

Answer 44

You're welcome! :D