2007-07-05 23:27:42 · 11 answers · asked by CPUcate 6 in Science & Mathematics ➔ Mathematics
(x-1 ) (x-2) (x+3) = 12
(x^2-3x+2)(x+3)=12
x^3-3x^2+2x+3x^2-9x+6-12=0
x^3-7x-6=0
X^3-x-6x-6=0
x(x^2-1)-6(x+1)=0
x(x-1)(x+1)-6(x+1)=0
(x+1)(x(x-1)-6)=0
(x+1)(x^2-x-6)=0
(x+1)(x^2-3x+2x-6)=0
(x+1)(x(x-3)+2(x-3))=0
(x+1)(x-3)(x+2)=0
x+1=0 or x-3=0 or x+2=0
x1=-1; x2=3; x3=-2
2007-07-05 23:46:52 · answer #1 · answered by cvet_che 2 · 4⤊ 0⤋
x=3
2007-07-14 01:55:01 · answer #2 · answered by JICKY 1 · 0⤊ 0⤋
x = 3
(x - 1 )(x - 2) (x + 3) = (3 - 1)(3 - 2)(3 + 3) = 2*1*6 = 12
2007-07-06 06:35:44 · answer #3 · answered by Northstar 7 · 0⤊ 2⤋
Answer: x must be 3.
Proof:
3 - 1 = 2
3 - 2 = 1
3 + 3 = 6
2 * 1 * 6 = 12
2 * 6 = 12
12 = 12
2007-07-13 00:53:49 · answer #4 · answered by Jun Agruda 7 · 2⤊ 1⤋
Learn how to use the "solve" function on a TI-89. Then plug and chug your possible answers.
2007-07-13 18:17:23 · answer #5 · answered by geezuskreyest 5 · 0⤊ 0⤋
the answer is 3
2007-07-12 01:02:31 · answer #6 · answered by p0oh 1 · 0⤊ 0⤋
(x - 1) (x² + x - 6) = 0
x³ + x² - 6x - x² - x + 6 = 0
x³ - 7x + 6 = 0
Using synthetic division shows that x - 1 is a factor:-
1|1***0***-7***6
*|****1****1***-6
---------------------
**1***1***-6***0
--------------------
(x - 1) (x² + x - 6) = 0
(x - 1) (x + 3) (x - 2) = 0
x = 1 , x = - 3 , x = 2
2007-07-11 03:52:59 · answer #7 · answered by Como 7 · 0⤊ 0⤋
(x-1 ) (x-2) (x+3) = 12
(x^2-3x+2)(x+3)=12
x^3-3x^2+2x+3x^2-9x+6-12=0
x^3-7x-6=0
X^3-x-6x-6=0
x(x^2-1)-6(x+1)=0
x(x-1)(x+1)-6(x+1)=0
(x+1)(x(x-1)-6)=0
(x+1)(x^2-x-6)=0
(x+1)(x^2-3x+2x-6)=0
(x+1)(x(x-3)+2(x-3))=0
(x+1)(x-3)(x+2)=0
x+1=0 or x-3=0 or x+2=0
x1=-1; x2=3; x3
2007-07-13 07:17:22 · answer #8 · answered by green 2 · 0⤊ 0⤋
3, cuberoot of 2
2007-07-06 06:34:16 · answer #9 · answered by Dominique 1 · 0⤊ 2⤋
(x-1)(x-2)(x+3)=12
x^3-7x+6=12
x^3-7x-6=0
x^2(x+1)-x(x+1)-6(x+1=0
(x+1)(x^2-x-6=0
x+1=0, x=-1
ORx^2-x-6=0
(x-3)(x+2)=0
x=3,-2
answer (-1,-2,3).
2007-07-06 09:42:47 · answer #10 · answered by Anonymous · 0⤊ 0⤋