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#### slowle4rner

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- Sep 2, 2013

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4y^3+4

It has been 7 years since I took algebra and I am trying to get my review done. This one just does not make sense to me right now...

Thanks!

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- Sep 2, 2013

- 2

4y^3+4

It has been 7 years since I took algebra and I am trying to get my review done. This one just does not make sense to me right now...

Thanks!

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- Sep 2, 2013

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4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)

That's basically what I came up with. I guess I am confused because I 4 is not a perfect cube and I didn't realize you could factor out the 4 and work the problem from there. But I suppose this makes sense assuming that I can multiply in the four and then multiply (4y+4)(y^2-y+1) or multiply 4(y+1)(y^2-y+1) and distribute the 4.

Would I be correct to presume that 4(y+1)(y^2-y+1) is the most factored and appropriate answer? That would seem to make sense to me...

Thanks for the help. I wasn't seeing it. Gratitude.

- Mar 1, 2012

- 249

The 4 is a bit of a red herring because, as you said, it's not a perfect cube but you can leave that alone out the front. You could say \(\displaystyle (4y+4)(y^2-y+1)\) (as well as \(\displaystyle (y+1)(4y^2-4y+4)\)) but it's good form to leave it fully factored which means not distributing the four. If you ever do exams they usually want it fully factored.So...\(\displaystyle a^3+b^3 = (a+b)(a^2-ab+b^2)\)

\(\displaystyle 4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)\)

That's basically what I came up with. I guess I am confused because I 4 is not a perfect cube and I didn't realize you could factor out the 4 and work the problem from there. But I suppose this makes sense assuming that I can multiply in the four and then multiply \(\displaystyle (4y+4)(y^2-y+1)\) or multiply \(\displaystyle 4(y+1)(y^2-y+1)\) and distribute the 4.

Yes, that is the correct answer.Would I be correct to presume that 4(y+1)(y^2-y+1) is the most factored and appropriate answer? That would seem to make sense to me...

Thanks for the help. I wasn't seeing it. Gratitude.

- Jan 29, 2012

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