I need help...
Where do I go from here?:
(x+1)(x-1)(x+1) = x(x+6)
*frustrated*
2006-09-18 10:49:19 · 7 answers · asked by Carly 2 in Science & Mathematics ➔ Mathematics
(x^2-1)(x+1)=x^2+6x
x^3+x^2-x-1-x^2-6x=0
x^3-7x-1=0
2006-09-18 10:56:45 · answer #1 · answered by raj 7 · 0⤊ 0⤋
Multiply the first two sets of parenthesis together gives
(x^2-x+x-1)(x+1) = x^2+6x
(-x+x) cancel each other out leaving
(x^2 - 1)(x + 1) = x^2 + 6x
x^3 + x^2 - x - 1 = x^2 + 6x
set equation to zero gives you
x^3 + x^2 - x^2 - x - 6x - 1 = 0
Simplify
x^3 - 7x - 1 = 0
2006-09-18 18:10:04 · answer #2 · answered by mitanbarr 3 · 0⤊ 0⤋
(x^2-1)(x+1) = x^2+6x
x^3-x+x^2-1 = x^2+6x
x^3 +x^2-x-1 = x^2 + 6
x^3 - x - 1 = 6
x^3-x = 7
x^3-x-7 = 0
2006-09-18 17:58:08 · answer #3 · answered by bob h 3 · 0⤊ 0⤋
Expand the brackets and collect terms:
X^3 -7x -1 = 0
It's a cubic equation. Can't be solved directly. Need to use newton raphson or a trial or computer method.
2006-09-18 17:52:24 · answer #4 · answered by Tammi J 3 · 1⤊ 0⤋
this is how you can go
(x^2-1)(x+1)=x^2+6x
x^3+x^2-x-1=x^2+6x
x^3 - 7x = 1
x(x^2-7) = 1
not much you can do now
2006-09-18 18:07:08 · answer #5 · answered by aursbe 2 · 0⤊ 0⤋
It is a cubic that can be solved. x=2.7
2006-09-18 18:06:27 · answer #6 · answered by Anonymous · 0⤊ 0⤋
thats easy but i dont want to waste my time on someone i dont even know .............sry........
2006-09-18 17:51:22 · answer #7 · answered by Liz 3 · 0⤊ 3⤋