Can you please factor this completly
2006-11-30 11:03:03 · 4 answers · asked by JOHN W 2 in Science & Mathematics ➔ Mathematics
(x-2)(x+1)(x-1)-6(x+1)
Now common factor is x+1
(x+1)[(x-2)(x-1)-6)]
(x+1)(x^2-3x+2-6)
(x+1)(x^2-3x-4)
(x+1)(x-4)(x+1)
(x-4)(x+1)^2
2006-11-30 11:10:52 · answer #1 · answered by Anonymous · 0⤊ 1⤋
(x-2)(x^2-1)-6x-6
First factor out a -6
(x-2)(x^2-1)-6(x+1)
-6(x-2)(x^2-1)x+1
Then expand
(x-2)(x^2-1) = -6(x^3-2x^2-x+2)x+1 Then multiply by x.
-6(x^4-2x^3-x^2+2x+1) now factor (x^4-2x^3-x^2+2x+1)
-6(x^2-x-1)^2 now expand (x^2-x-1)^2
-6(x^2-x-1)(x^2-x-1)
2006-11-30 11:53:11 · answer #2 · answered by Pam 5 · 0⤊ 1⤋
(x-2)(x-1)(x+1) - 6(x+1) = (x+1)[(x-2)(x-1)-6]
Factor only the part within the [ ]:
x^2-3x+2-6 = x^2-3x-4 = (x-4)(x+1), so that we have...
(x+1)[(x-4)(x+1)] = [(x+1)^2](x-4)
2006-11-30 11:11:03 · answer #3 · answered by kellenraid 6 · 0⤊ 1⤋
multiply
(x-2)(x^2-1) = x^3-2x^2-x+2 - 6x - 6 = then just combine like terms
2006-11-30 11:06:42 · answer #4 · answered by Ambino 4 · 1⤊ 1⤋