Factor the GCF and then the resulting trinomial: 3x^3 - 3x^2 - 6x
Teach dont give me the answer

- anonymous

Factor the GCF and then the resulting trinomial: 3x^3 - 3x^2 - 6x
Teach dont give me the answer

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- anonymous

Good for you, jonnyboy66 - I'm impressed that you don't want the answer! That's great! :) You'll go far!

- anonymous

Okay, now take a look at what is common for EACH TERM, in the trinomial. What numbers and x's do you see that each term has in common, that could be taken and put outside of parentheses?

- anonymous

im not sure

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- anonymous

I really wanna understand though

- anonymous

there are many 3's and 2 x's

- anonymous

Okay, good! That's a good start! Well, let's take a look at what we've got . . . .

- anonymous

Are you there?

- anonymous

I gotta go thanks for your help though

- anonymous

can you message me what you were gonna say

- anonymous

and Ill look at it

- anonymous

When you factor the GFC, remember that the GFC stands for "Greatest COMMON Factor." What that means is - you're looking to take out of the equation (and put outside of parentheses), numbers and variables (remember - variables are letters, like "x" and "y") that EACH term has in common with ALL of the other terms.
Okay, now when you look at the expression (3x^3 - 3x^2 - 6x), take a look at each term (a term can be just a number - like 3 - or have variables - like xy. Basically, a term is like a group of something). What do you see that each term has, that ALL of the other terms have? Well, if we look, we can see that each term has a 3 (remember that 3 is a multiple of 6). So, we know that the expression has a 3 in common. Next, we need to look at the variables (the letters), to see if each term has something that all of the other terms have. Now, when you go to see how many x's all of the terms have in common, you MUST only look at the variable with the LEAST amount of x's. The reason for this is . . . .

- anonymous

that when we go to take an "x" out of the equation, whatever we take out, must, when multiplied back into the equation, make each of the terms go back to what they were, originally (don't worry if this sounds confusing - you'll understand in a minute, what I'm meaning). Okay, so, look at the expression. See how the lowest "x" in the equation is the "x" for the term (6x)? Well, that means that we can only take 1 "x" out of the equation. If we were to take more than that, when we multiplied the "x's" outside of the brackets back into the original expression, we would end up with that "x" in the (6x) being raised to a much higher power, which is NOT the original equation that we have, here. Okay, so now that we can see that a 3 and an "x" are in common for EACH term in this expression. We need to factor out (or take out of the expression, put the expression inside of parentheses, and then put the 3 and the "x" outside of the parentheses) the 3 and the "x" from the expression. So, here's the what the setup looks like: 3x( ) Inside of those brackets, we're going to put our NEW expression, which will now have 1 fewer 3 and 1 fewer "x" for EACH term in the equation (because we took a 3 and an "x" out of the equation). Now for the next step . . . . .

- anonymous

Now that we have a (3x) taken out of the expression, we're left with a new expression, where each term will be missing one 3 and one "x". I'll show you what the new expression looks like: \[3x (x^2 - x - 2)\] Notice the expression inside of the parentheses. It's now become: \[(x^2 - x - 2)\] What happened to the variables ("x" and "y") and coefficients (or numbers in front of the "x's" and "y's") of 3 and 6? Well, what you've done, is take out a 3 and an "x" from each term. Let's look at each term, to see what happened . . . .

- anonymous

So, here's the original expression: \[(3x^3 - 3x^2 - 6x)\] Now, when we took a 3 from the 3x^3 (the first term in the expression), the 3 disappears, leaving us with: x^3 Then, when we take one x from the x^3, it becomes the x^2. Now for the second term (the 3x^2). When we take a 3 from that, it leaves us with: x^2. Then, when we take 1 "x" from that, we're left with: x. And now for the third term (6x). When we take a 3 from the 6, we're left with 3x (remember that there are two 3's in a 6 - 3 x 2 = 6 - when we took one 3 away, we were left with 1 three). Then, when we take an x from the 3x, that then leaves us with 3 (the "x" completely disappears from the 3 because we've taken the only "x" that it has).
Okay, now I hope this isn't too confusing. Our last step in factoring out the GCF is to take the GCF (the 3 and the "x") and put it outside of parentheses.
So, here's what we have: \[3x (x^2 - x - 2)\]
So . . . after all of that hard work, we've finally factored out the GCF. But .. . . we're not done, yet. We now need to factor the expression in the parentheses. Phew! - hang in there - we're almost done! :)

- anonymous

Okay, now in order to factor, we're going to set the stuff in brackets (the (x^2 - x - 2)) into 2 separate parentheses. It will kind of look like this: (x )(x ) Note: we don't yet know if we will be adding or subtracting numbers, so that's why there's nothing there. Okay, now if you remember, we need to find 2 numbers that will ADD together to create the MIDDLE term in our equation (that would be the invisible coefficient of -1, which is in front of the "x"), and 2 numbers that will MULTIPLY together to create the very last term (that would be the -2). Note: the negatives are there because, in the expression, they're being subtracted. Now, let's think about this: well, if we ADD the numbers 1 and (-2) together, we get our middle term of (-1), and if we MULTIPLY the numbers 1 and (-2) together, we get our last term (in the expression) of (-2). So, the numbers that we need to use for the parentheses are: (-2),1. Okay - let's plug them in: (x-2)(x+1) Note: the reason why there is a minus sign in front of the 2 is because the negative sign in front of the (-2), showed us that we will be SUBTRACTING 2 from "x".
So . . . . here's you're final and factored answer: 3x(x-2)(x+1)
Note: remember to put the 3x back in front of the (x-2) and the (x+1) - you still keep that in front.
I hope this helped! Sorry it took so long. Good luck! :)

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