Here's what I've done so far:
(8+x^3)(8-x^3)
(2+x)(4-2x+x^2)(2-x)(4+2x+x^2)
Can it be further factored?
Thanks all in advance!
2007-09-14 19:26:44 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Yup that is enough. You cannot further factor (4-2x+x^2) since it has no longer has real roots. Try to verify this by applying the quadratic equation.
I hope I answered your question.
2007-09-14 19:40:58 · answer #1 · answered by Enrico C 2 · 0⤊ 0⤋
x^6 - 64 = (x^3 -8) (x^ +8) it's basically the same problem as y^2 - 64 where y = x^3 So the answers are 2 and -2
2016-05-20 00:09:09 · answer #2 · answered by ? 3 · 0⤊ 0⤋
64 - x^6
= [8² - (x³)²]
= (8 - x³)(8 + x³)
= (2 - x)(4 + 2x + x²)(2 + x)(4 - 2x + x²)
2007-09-14 20:40:03 · answer #3 · answered by Northstar 7 · 0⤊ 0⤋
= 64 - x^6
= (8 + x^3) (8 - x^3)
I think these are primes and cannot be factored further.
2007-09-14 19:42:21 · answer #4 · answered by Jun Agruda 7 · 3⤊ 1⤋
(2^3 + x^3) (2^3 - x^3)
you just have to do it until here. no need to expand any further.
2007-09-15 04:15:15 · answer #5 · answered by Anonymous · 0⤊ 0⤋
(2+x)(4-2x+x^2)(2-x)(4+2x+x^2)
= (2+x)((x-1)^2+4-1)(2-x)((x+1)^2+4-1)
= (2+x)((x-1)^2+3)(2-x)((x+1)^2+3)
2007-09-14 19:37:20 · answer #6 · answered by JoZZ 2 · 0⤊ 0⤋
64 = 2^6
so x^6 = 2^6
factors will be (x-2)(x-2)(x-2)(x-2)(x-2)(x-2)
you can make any combinations of it
2007-09-14 19:40:12 · answer #7 · answered by libraboy28 2 · 0⤊ 1⤋
answer:
-(x-2)(x+2)(x^2-2x+4)(x^2+2x+4)
2007-09-14 19:35:01 · answer #8 · answered by kimbokrn 2 · 0⤊ 1⤋
No.
What you have done is correct.
2007-09-14 19:38:54 · answer #9 · answered by Madhukar 7 · 0⤊ 0⤋