Alright, so I have $x^3 + 2x^2y - 4x - 8y$. I've learned that I need to group them together, so I chose to group $2x^2y - 4x$ and $x^3 - 8y$

When taking out their GCF, the numbers left in the parenthesis didn't match up, in fact, I couldn't find the GCF of anything for the second group.

So there is either a solution or it is prime, according to my algebra teacher, however, I'm not sure how to tell if something is prime or not. Can someone please help me?

EDIT: Could it also be that I am just grouping them incorrectly?

SECOND EDIT: Alright, I regrouped them a different way, into (-4x + x^3) + (2x^2y - 8y). After factoring them out, I came up with x(-4 + x^2) + 2y(x^2 - 4). The terms in the parenthesis are the same, but they are flipped. Does this mean that my answer is incorrect?

  • $\begingroup$ Try grouping the terms the correct way now. $\endgroup$ – whacka Feb 19 '15 at 1:51
  • $\begingroup$ How do I do it the "correct way"? I've been told that the grouping doesn't matter, so long as you fully factor the terms. $\endgroup$ – johny Feb 19 '15 at 1:52
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    $\begingroup$ Remarkably unhelpful, @whacka $\endgroup$ – Thomas Andrews Feb 19 '15 at 1:53
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    $\begingroup$ Regarding your second edit: Addition is commutative. $\endgroup$ – Thomas Andrews Feb 19 '15 at 1:57
  • $\begingroup$ @Thomas: I made my comment before the (first) edit. OP: there are only so many ways to try out, so it can't hurt to try them all! You don't always start out a problem knowing the correct way to do it, which is why trying is important. And as Thomas points out, $-4+x^2$ and $x^2-4$ are the same thing, so you might as well write the same thing in both parentheses: $x(x^2-4)+2y(x^2-4)$. (How did you end up moving the $-4x$ term before the $x^3$ term anyway?) $\endgroup$ – whacka Feb 19 '15 at 1:57

$$x^3 + 2x^2y - 4x - 8y = (x^3 + 2x^2y) - (4x + 8y) = x^2(x+2y) - 4(x+2y) = (x^2-4)(x+2y)$$ $$\cdots = (x-2)(x+2)(x+2y)$$


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