2007-05-14 15:43:33 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The more obvious common factor here is that 128 is divisible by 8, 16 times.
So, pull out the 8, and you get: 8(16x^2-1).
Now, 16x^2-1 happens to factor into (4x+1)(4x-1). You can usually tell when you have a something that factors like this when you have only an x^2 and a negative constant, and both the constant and the coefficient have rational sqaure roots.
So, now you have 8(4x+1)(4x-1)
2007-05-14 15:50:30 · answer #1 · answered by Anonymous · 0⤊ 0⤋
128x^2-8
=8(16x^2-1)
=8{(4x)^2-(1)^2}
=8(4x+1)(4x-1) ans.
2007-05-14 16:46:06 · answer #2 · answered by alpha 7 · 0⤊ 0⤋
8 (16 x^2 - 1)
8(4x -1) (4x + 1)
2007-05-14 15:49:27 · answer #3 · answered by f4llen4ngel 2 · 0⤊ 0⤋
8(16x² - 1)
= 8(4x-1)(4x+1)
2007-05-14 15:52:38 · answer #4 · answered by Kathleen K 7 · 0⤊ 0⤋
= 8.(16x² - 1)
= 8.(4x - 1).(4x + 1)
2007-05-14 21:00:00 · answer #5 · answered by Como 7 · 0⤊ 0⤋