Question

Factor completely
1. a^3-ab+a^2-b
2. g^5-g
3. x^3-27y^9
4. 64x^6+8n^3 Factor completely
1. a^3-ab+a^2-b
2. g^5-g
3. x^3-27y^9
4. 64x^6+8n^3 @Mathematics

Answer 1

1.(a^2-b)(a+1)
2.g(g^4-1)
3.(x-3y^3)(x^2+3xy^3+y^6) (difference of cubes)
4.(4x^2+2n)(16x^4-8x^2n+4n^2) (sum of cubes)

Answer 2

g(g^4-1) - Can't you do the difference of 2 perfect squares? g(g^2-1)(g^2+1) ?

Answer 3

yes, my bad I spaced out on that one :)
you can actually do it twice:
3.g(g-1)(g+1)(g^2+1)

Answer 4

And for this: (4x^2+2n)(16x^4-8x^2n+4n^2) when you put -8x^2n is that what you meant?

Answer 5

yes, sum of cubes is
a^3+b^3=(a+b)(a^2-ab+b^2)
here you have
a=4x^2
b=2n

Answer 6

Wouldn't (a^2-ab+b^2)
= (16x^4-(4x^2)(2n)+4n^2)?

Answer 7

Not: (16x^4-8x^2n+4n^2)?

Answer 8

yes, which simplifies to
16x^4-8x^2n+4n^2
(4x^2)(2n)=8x^2n
you could put parentheses I guess if it looks confusing
16x^4-(8x^2)n+4n^2

Answer 9

Oh I got it, I thought you raised it to the power of 2n, its confusing on the computer

Answer 10

Thanks, do you think you're up for 1 more?

15x^2-43x-380

Answer 11

15x^2-43x-380
(3x-20)(5x+19)
i cheated for that one, it looked annoying.
http://www.wolframalpha.com/input/?i=factor+15x^2-43x-380

Answer 12

Wolfram can do most of those.

Answer 13

Thanks so much, you're the best. I never heard of that and it will save me lots of time. :)