SOLUTION: i need to factor this but i got stuck! so here is what i already have: x^3<x^2+9x-9 = x^3-x^2-9x+9<0 = ?

Algebra ->  Equations -> SOLUTION: i need to factor this but i got stuck! so here is what i already have: x^3<x^2+9x-9 = x^3-x^2-9x+9<0 = ?       Log On


   

Question 49418: i need to factor this but i got stuck! so here is what i already have:
x^3
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
x^3 - x^2 - 9x + 9 < 0
a = 1,3,9
b = 1
Possible Roots: +- a/b = +-1,3,9
Descarte's Rule of Signs:
f(x) = (+) ~> (-) ~> (-) ~> (+)
Pos. = 2
f(-x) = (-) ~> (-) ~> (+) ~> (+)
Neg. = 1
Imag. = 0
so:
pos. = 2 or 0
neg. = 1 or 1
imag = 0 or 2
There is a definately a root that is negative.
Synthetic Division (if you do not know how to do this, just divide easily):
-3|....1....-1....-9....9
.........1....-4.....3.....0
(x + 3)(x^2 - 4x + 3) < 0
(x + 3)(x - 3)(x - 1) < 0
Since this data is less than zero, you want the negative part of the graphed line in a cartesian coordinate plane.
-4 ~> negative
-3 ~> zero
0 ~> positive
1 ~> zero
2 ~> negative
3 ~> zero
4 ~> positive
<====(-3)----(1)====(3)---->
(-infinity,-3)U(1,3)
graph%28300%2C300%2C-5%2C5%2C-6%2C6%2Cx%5E3-x%5E2-9x%2B9%2C0%29