Question

Factor x^12 – 64 completely.

A. (x^6 – 8)(x^6 + 8)

B. (x^4 – 4)(x^8 + 4x^4 + 8)

C. (x^3 – 2)(x^3 + 2)(x^6 + 2x^3 + 4)(x^6 – 2x^3 + 4)

D. (x^2 – 2)(x^2 + 2)(x^4 + 2x^2 + 4)(x^4 – 2x^2 + 4)

Answer 1

Help!:(

Answer 2

First, remember that a difference of two squares can be factored like this:

a² - b² = (a + b)(a - b)

Now, because \(x^{12}=(x^6)^2\) and \(64=8^2\), we have the difference of two squares here!

So you can factor it as I described.

Answer 3

ok, thanks

Answer 4

the answer is D if you factor this completely
1. start with (x^6)^2 - (2^3)^2
2. follow the rule: A^2 - B^2 = (A-B)(A+B)
your A will be (x^6) and your B will be (2^3)
3. so, it goes this way:
(x^6 - 2^3)(x^6+2^3)
4. factor x^6
[(x^2)^3 - 2^3] [(x^2)^3 + 2^3]
5. now, as you can see, there are two parts enclosed each one with braces [ ]
6. in the first part, you follow the rule: A^3 - B^3 = (A-B)(A^2+AB+B^2)
and in the second part, follow this rule: A^3 + B^3 = (A+B)(A^2-AB+B^2)
wherein A = x^2 and B = 2
7. substitute your values of A and B in the right side of each rule
8. you can now simplify each part
9. can you now notice the similarity of the expression to the choices?
10. if not, arrange it properly to see your final answer and this can be found in letter D... :)

Answer 5

Thank you! A LOT!