Question

81a^3-4a

Answer 1

What about it?

Answer 2

trinomial factorization

Answer 3

Ok, well, this is a binomial (or you can think of it as a trinomial where the middle term = 0).
Factor out the GCF of both terms, then you'll have a difference of squares.

Answer 4

ok so what will it look like

Answer 5

(9a^2-9a)(2-2a)

Answer 6

im so confused

Answer 7

Er .. .
Ok, with those two terms, the only factor they both have in common is an a, so take that out and what do you have left?

Answer 8

81-4

Answer 9

The first term had three a's, so when you take one out, it still has a^2 left.

Answer 10

81^2-4...then what do i do

Answer 11

i mean 81a^2

Answer 12

That is now a difference of squares. There is a pattern that always works for a difference of squares. You can do it the usual way for a quadratic trinomial understanding that the middle term is zero.

Answer 13

so would it be (9a-2)(9a+2)

Answer 14

no thats not right nvm

Answer 15

Yes, once you see that\[81a^2-4 = (9a)^2-(2)^2.\] then the factorization is easy. Just don't forget to include that a that was factored out at the beginning is still part of the product, so include it with the other factors.

Answer 16

ok so my final product should look like what

Answer 17

is the final answer (9a^2+2a)(9a-2)

Answer 18

mmm, that's true, but leave that a that was common to both terms at the beginning on the outside as a separate factor: (a)(9a+2)(9a-2)

Answer 19

If you distribute it back in then you'll undo the factoring that you already did.

Answer 20

ok so what doi need to do? what am i doing wrong?

Answer 21

ok so a(9a+2)(9a-2) is the answer

Answer 22

You can always check your answer to factorization by multiplying all the terms back together to see if you get back what you started with.

Answer 23

i did and it worked but it just seems like there's more too it