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

In summary, the problem is to factor the binomial 4y^3+4. By factoring out the 4, we get 4(y^3+1). Applying the sum of cubes formula, we get 4(y+1)(y^2-y+1) as the most factored and appropriate answer. It is good form to leave it fully factored, without distributing the 4. The "4" in the beginning is just a number and can be factored out. Thank you for the help!
  • #1
slowle4rner
2
0
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!
 
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  • #2
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?
 
  • #3
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.
 
  • #4
slowle4rner said:
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.
 
  • #5
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.
 

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

1. How do you factor a binomial with a common factor?

To factor a binomial with a common factor, you need to find the greatest common factor (GCF) of the two terms. In this case, the GCF is 4y^3. You can then divide each term by the GCF and write the expression as the product of the GCF and the remaining terms, giving you 4y^3(1+1).

2. Can a binomial with a common factor be further factored?

Yes, a binomial with a common factor can be further factored. In the example given, the expression 4y^3(1+1) can be simplified to 4y^3(2), which can then be factored to 4y^3(2)(1+0). This gives you the final factored form of 8y^3(1+0).

3. How do you factor a binomial with a coefficient?

To factor a binomial with a coefficient, you can follow the same steps as factoring a binomial with a common factor. First, find the GCF of the two terms. Then, divide each term by the GCF and write the expression as the product of the GCF and the remaining terms. In this case, the GCF is 4, so the factored form would be 4(1+y^3).

4. Can a binomial with a coefficient be further factored?

Yes, a binomial with a coefficient can be further factored. In the example given, the expression 4(1+y^3) can be simplified to 4(1+y^3)(1). This gives you the final factored form of 4(1+y^3)(1).

5. How do you check if a factored binomial is correct?

To check if a factored binomial is correct, you can use the distributive property. Simply multiply the terms in the factored form together and see if you get the original expression. In the example given, multiplying 4y^3(1+1) gives you the original expression of 4y^3+4. This confirms that the factored form is correct.

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