Question

hey dose anyone know how to do the problem 2y^3 - 128 the formula for this that I am trying to use is difference of squares A^3 - B^3 = (A-B) (A^2 + AB + B^2)

the direct5ions fo this problem are to factor out completely i am very confused please help!

Answer 1

first off, factor 2 out, you have 2(y^3-64) , ok now, right?

Answer 2

yes

Answer 3

good!

Answer 4

okay so whats the next step and your getting the 64 from 128 right?

Answer 5

could you use the draw thing if you could show your steps that would help a lot.

Answer 6

hint: 64 = 4^3

so 2(y^3-64) = 2(y^3-4^3)

Answer 7

:) @huntergirl1 When you said yes, I thought you got all of them since you gave me the formula.
You just apply the formula you post to get the answer.

Answer 8

okay so i end up with what i started with which is great because that was the correct answer and yes sorry i was having a dyslexia moment with the 64 thing i understand what you got and this is what i got...


2(y-4) (y^2 +4y +16)
2(y^3 + 4y^2 +16y -4y^2 -16y-64) i simplied with like terms and then i got this.
2(y^3-64)= 2y^3-128 the original problem

Answer 9

unless i am completely wrong

Answer 10

2y^3 - 128

2(y^3 - 64)

2(y^3 - 4^3)

2(y-4)(y^2 + y*4 + 4^2) ... apply the difference of cubes rule here

2(y-4)(y^2 + 4y + 16)

and that's as far as you can go

Answer 11

so yes this is correct then?

Answer 12

the steps after 2(y-4)(y^2 + 4y + 16) appear to be you expanding everything out again, which is the opposite of what they want. They want you to factor.

Answer 13

okay then so all i need is the set up part?

Answer 14

what do you mean?

Answer 15

so im leaving it at 2(y-4) (y^2 +4y +16)

Answer 16

or is that not fully factored out?

Answer 17

yeah that's the fully factored form

Answer 18

okay thanks!! (:

and also would you be up to helping me with one more?

rs^2 +64?

same directions and would i start by factoring the r out completely?

Answer 19

I don't see an 'r' in the second term

Answer 20

is there one?

Answer 21

sorry typo that 64 is supposed to have an r on it as well
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Answer 22

rs^3 + 64r

should i take and factor out the r first?

Answer 23

yes, and what do you get when you do so?

Answer 24

r(s^3 +4^3)

Answer 25

is that on the right track?

Answer 26

yes it is

Answer 27

then you'll use the sum of cubes factoring formula

Answer 28

and once it is plugged into the formula then i am done right? with the factoring?

Answer 29

yes, tell me what you get

Answer 30

okay i got r(s+4) (s^2- 4s+ 16)

Answer 31

correct

Answer 32

okay and then i went on and i did this r(s^3 - 4s^2 + 16s+ 4s^2 -16s + 64) I'm going to combine like terms and then end up with...

r(s^3 + 64) right or did i go to far

Answer 33

why are you expanding again? they just want you to factor

Answer 34

r(s+4) (s^2- 4s+ 16) is the fully factored form

when you got to r(s^3 - 4s^2 + 16s+ 4s^2 -16s + 64), it's correct but not what they want. They don't want you to expand. I'm assuming they simply said "factor the expression"

Answer 35

they just said fator ompletely but okay so then leave it at that

Answer 36

yeah, r(s^3 - 4s^2 + 16s+ 4s^2 -16s + 64) takes a step backward from factoring

so leave it as r(s+4) (s^2- 4s+ 16)

Answer 37

okay then but if i was going to check this because i have a quiz on this information tomorrow the I would continue to expand? i would show my answer as-->r(s+4) (s^2- 4s+ 16) then expand to check my answr to make sure that it matches the original problem right? .

Answer 38

that is correct

Answer 39

okay cool!! thank you soooo much!!! i will not torture you anymore with my questions thanks for the help!!

Answer 40

lol you're ok, no worries. I'm glad I could help out