Question

I just need help with part B! I think I know what to do, but I need help..
The GCF factored out is x^3+4x -2x^2 -8 I hope
An expression is shown below:

2x^3y + 8xy − 4x^2y − 16y

Part A: Rewrite the expression so that the GCF is factored completely. (4 points)

Part B: Rewrite the expression completely factored. Show the steps of your work. (6 points)

Answer 1

A. In GCF, "CF" is Common Factor. -> x^3+4x -2x^2 -8 is not the GCF, it's the other factor: y. It is not posible to factor y more than 'just y'.
But you must have this product right: \(2x^3y + 8xy − 4x^2y − 16y = (x^3+4x -2x^2 -8)y\).

B. To factor \(x^3+4x -2x^2 -8\), you must find one of its roots: try 0, 1, -1, 2, -2 .. until it's right.

Answer 2

Oh, alright then.. So then how would I factor out the GCF correctly?

Answer 3

with 1 : 1^3 + 4 * 1 - 2 * 1^2 - 8 ≠ 0
with -1 : (-1)^3 + 4 * (-1) - 2 * (-1)^2 - 8 ≠ 0
with 2 : 2^3 + 4 * 2 - 2 * 2^2 - 8 = 0 yay
Do you know what to do now?

Answer 4

The GCF (greatest common factor) is 'y', and it is already completely factored out. Can't write it as a product of several things.

Answer 5

I'm sorry... I am so confused. Did I do the first part correctly, or not? Oh wait, I meant the GCF is 2y im pretty sure. What I have up there is it factored out.. Oh snap, I have no idea what I am doing.

Answer 6

\( 2x^3y + 8xy − 4x^2y − 16y = (x^3+4x -2x^2 -8)(2y) \)
The GCF is \(2y\). To factor it out completely, we just write \(2y = 2 \times y\)...
(I forgot the '2' above in an earlier reply)

Answer 7

Okay, thank you, I understand that part. Now, how exactly would I factor for part B? I thought maybe I would just factor by grouping, but maybe not. I suck at algebra, as you can see.

Answer 8

That's it for part A.

For part B, grouping seems to work as "\(x^2 +4\)" is going to show up several times.

Answer 9

Okay, thank you! I think I can finish it now.

Answer 10

nice, good luck

Answer 11

\(2\times y \times (something) \times (something)\).
I count 4 factors in the final factorization.

Answer 12

Oh, okay. Sorry, but could you help me factor it completely for part b? Im sure I am doing it wrong.

Answer 13

And I apologize, my computer is super slow...

Answer 14

Oh okay, you're a genius, I swear. And Im looking at the first group, and something looks wrong, maybe not. Im bad at math.

Answer 15

\(\underbrace{x^3 + 4x}_{x(x^2+4)} \underbrace{- 2x^2 - 8}_{-2(x^2+4)} = (x^2+4)\bigl( x -2 \bigr)\)

Answer 16

I messed up with the exponenets.. so i reposted the correct version.

Answer 17

Okay, thank you :) I was wondering about that. And thank you for all your help! It is appreciated greatly.