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help factoring

slowle4rner

New member
Sep 2, 2013
2
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Sep 2, 2013
2
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
Mar 1, 2012
249
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
Jan 29, 2012
1,151
Even if you were not able to factor that "4" out, it just a number! If it had been, say, [tex]x^3+ 4[/tex] you could write it as [tex]x^3+(\sqrt[3]{4})^3[/tex] and use that same fornula.