Question

factor completely

a^4c^4 - 81

Answer 1

meant a^4x^4- 81

Answer 2

Do you recognize this as the differrence of 2 squares?

Answer 3

a^2 - b^2 = (a + b)(a - b)

Answer 4

yes

Answer 5

a^4x^4 - 81 = (a^2x^2 - 9)(a^2x^2 + 9)
can it be factored further - i'm looking at the first bracket

Answer 6

no

Answer 7

Hint: 9 = 3^2

Answer 8

Are you with me so far?

Answer 9

yes

Answer 10

Correction:
\[a^4x^4-81=[(a^2x^2)^2-9^2]=(a^2x^2-9)(a^2x^2+9)\]

Answer 11

\[a ^{2}x ^{2} = (ax)^{2} and 9 = 3^{2}\]

Answer 12

Does that help?

Answer 13

So now please note that \[a^2x^2-9 \] is also the difference of two squares and can be factored again.

Answer 14

alright

Answer 15

\[a^2x^2-9=(ax-3)(ax+3)\]

Answer 16

Still with me?

Answer 17

yes

Answer 18

That's it, that's the answer:
\[(ax + 3)(ax + 3)(a ^{2}x ^{2} + 9) \]

Answer 19

So we now have the following for the completely factored form:
\[(ax-3)(ax+3)(a^2x^2+9)\]

Answer 20

Thank you!

Answer 21

And, of course, the answer pratu posted is incorrect.

Answer 22

yw. Do you understand?

Answer 23

can you please give the best answer thing to pratu043 since i cant ?

Answer 24

Sorry, its ax - 3

Answer 25

alright & yes i understand thanks!

Answer 26

You don't need to I gave the wrong answer.

Answer 27

Thanks anyway.