2007-03-29 16:17:58 · 9 answers · asked by Angie D 1 in Science & Mathematics ➔ Mathematics
32x^3 - 4
4(8x^3 - 1)
4((2x)^3 - 1^3)
4(2x - 1)((2x)^2 + 2x + 1)
4(2x - 1)(4x^2 + 2x + 1)
2007-03-29 16:21:45 · answer #1 · answered by Anonymous · 0⤊ 0⤋
You need to find the greatest common factor that goes into both 32 and 4, which is 4.
so the answer is
4(8^3-1)
to check your answer: if you distribute you get 32x^3-4
2007-03-29 16:41:16 · answer #2 · answered by Justina 3 · 0⤊ 1⤋
It goes like this:
32x^3 - 4 = 4 (8x^3 - 1) *divide every term by 4*
2007-03-29 16:21:58 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1) common factor 4:
4(8x^3 - 1)
This is a difference of cubes
8x^3 = (2x)^2 and 1 = 1^3
2) Next step, factor out (2x - 1)
4(2x - 1)(4x^2 + 2x + 1)
The quadratic does not have roots in real numbers, so you stop here.
2007-03-29 16:24:55 · answer #4 · answered by Raymond 7 · 0⤊ 0⤋
take out 4
4(8x^3 - 1)
differrence of perfect cubes
a^3 - b^3 = (a-b) (a^2 + ab + b^2)
2^3x^3 - 1^3
(2x)^3 - 1^3
4 (2x - 1) (4x^2 + 2x + 1)
2007-03-29 16:23:45 · answer #5 · answered by 7 · 0⤊ 0⤋
take out a 4
4(8x^3 -1)
difference of two cubes
a = 2x
b = 1
(2x-1)(4x^2+2x+1)
gotta memorize those difference/sum of two cube formulas
dif:
(a^3-b^3) = (a-b)(a^2+ab+b^2)
(a^3 + b^3) = (a+b)(a^2 - ab +b^2)
2007-03-29 16:23:53 · answer #6 · answered by yup5 2 · 1⤊ 0⤋
GCF:4
4(8X^3-1)
2007-03-29 16:21:26 · answer #7 · answered by :) 5 · 0⤊ 1⤋
ok initiate right here x circumstances x = x^2 it somewhat is x squared or x circumstances itself so x(x+a million) could equivalent x^2+x Now 2X(x+a million) could equivalent 2x^2+2x so now 2x(3x-4) could equivalent 6x^2-8x Hopes it facilitates
2016-12-08 14:14:43 · answer #8 · answered by ? 4 · 0⤊ 0⤋
use your computer...Duh !!! don't apply to NASA !
2007-03-29 16:20:31 · answer #9 · answered by Anonymous · 0⤊ 4⤋