Solve for x.
2007-08-23 19:47:08 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^3 - 2x^2 - 5x + 6
= x^3 + 2x^2 - 4x^2 - 8x + 3x + 6
= x^2 (x + 2) - 4x(x +2) + 3 (x + 2)
= (x + 2) (x^2 - 4x + 3)
= (x + 2) (x^2 - x - 3x + 3)
= (x + 2) [x (x - 1) - 3 (x - 1)]
= (x + 2) (x - 1) (x - 3)
= (x - 1) (x + 2) (x - 3)
So x = 1, -2, 3
2007-08-23 19:48:49 · answer #1 · answered by Sam 3 · 1⤊ 1⤋
Factor by grouping.
(x^2-4x+3)(x+2).
Factor again:
(x-1)(x-3)(x+2) = 0
So x is 1, 3, & -2.
2007-08-24 02:55:43 · answer #2 · answered by SoulDawg 4 UGA 6 · 0⤊ 0⤋
(x-3)(x^2 +x -2) = 0
(x-3) (x-1) (x+2) = 0
x = 3 , 1 , -2
2007-08-24 02:57:17 · answer #3 · answered by CPUcate 6 · 0⤊ 0⤋
check x=1 is a solution
so write
(x-1)(x²-x-6)=0
and you have a simple binomial equation left
2007-08-24 02:52:17 · answer #4 · answered by Champoleon 5 · 0⤊ 1⤋