How do I factor a binomial with a coefficient of 4?

[slowle4rner]

I'm having trouble factoring the following binomial... can someone try to point me in the right direction please?

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!

 

[MarkFL]

First factor the 4 out, to get:

$$4y^3+4=4\left(y^3+1 \right)=4\left(y^3+1^3 \right)$$

Now apply the sum of cubes formula:

$$a^3+b^3=(a+b)\left(a^2-ab+b^2 \right)$$

What do you get?

 

[slowle4rner]

So...A^3+b^3 = (a+b)(a^2-ab+b^2)

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.

 

[SuperSonic4]

So...[math]a^3+b^3 = (a+b)(a^2-ab+b^2)[/math]

[math]4y^3+4 = 4(y^3+1) = 4(y^3+1^3) = 4(y+1)(y^2-y+1)[/math]

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.

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 $$(4y+4)(y^2-y+1)$$ (as well as $$(y+1)(4y^2-4y+4)$$) 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.

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.

Yes, that is the correct answer.

 

[HallsofIvy]

Even if you were not able to factor that "4" out, it just a number! If it had been, say, $$x^3+ 4$$ you could write it as $$x^3+(\sqrt[3]{4})^3$$ and use that same fornula.