# help factoring

#### slowle4rner

##### New member
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

Staff member
First factor the 4 out, to get:

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

Now apply the sum of cubes formula:

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

What do you get?

#### slowle4rner

##### New member
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

##### Well-known member
MHB Math Helper
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.
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.

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

##### Well-known member
MHB Math Helper
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.