2007-12-29 06:41:11 · 7 answers · asked by ppd232 1 in Science & Mathematics ➔ Mathematics
4x³ - 4x² + 28x - 28
4(x³ - x² + 7x - 7)
4(x - 1)(x² + 7)
These factors are irreducible over R, but if you don't mind using complex numbers, we can do one more step:
4(x - 1)(x + i√7)(x - i√7)
2007-12-29 06:45:30 · answer #1 · answered by Pascal 7 · 0⤊ 0⤋
Group first 2 terms together, then the last 2 terms together:
(4x^3- 4x^2) + (28x - 28)
Factor out a GCF for each separate binomial:
4x^2(x-1) + 28(x-1)
Factor out the common binomial:
(4x^2 + 28)(x-1) = 4(x^2 + 7)(x-1)
2007-12-29 06:55:41 · answer #2 · answered by elin1081 2 · 0⤊ 0⤋
4x^3 - 4x^2 + 28x - 28
=>4(x^3 - x^2 + 7x - 7)
=>4[x^2(x-1) + 7(x-1)]
=>4(x-1)(x^2+7)
2007-12-29 06:47:18 · answer #3 · answered by mohanrao d 7 · 0⤊ 0⤋
4x^3 - 4x^2 + 28x - 28
4(x^3 - x^2 + 7x - 7)
4[x^2(x-1)+7(x-1)]
4[(x^2 + 7) (x-1]
2007-12-29 06:47:24 · answer #4 · answered by Namu 2 · 0⤊ 0⤋
4x² - 4x - 3 = 0 the classic technique of (4x )(x ) won't artwork because of the fact the midsection term would be off no remember how we place the 'a million' and the '3' and the '+'s and '-'s. So, the appropriate we are in a position to do is to tug out 4x 4x(x - a million) - 3 = 0
2016-12-18 11:20:23 · answer #5 · answered by ? 4 · 0⤊ 0⤋
4x^2(x-1) +28(x-1)
= (4x^2+28)(x-1)
= 4(x^2+7)(x-1)
2007-12-29 06:45:41 · answer #6 · answered by ironduke8159 7 · 0⤊ 0⤋
p = 4·(x³-x²+7x-7) = 4·(x-1)·(x²+7)
saludos.
2007-12-29 06:46:40 · answer #7 · answered by lou h 7 · 0⤊ 0⤋