Posted by anon on Saturday, December 10, 2011 at 10:27pm.
I think you meant
10x^3 + 6x^2 - 90x - 54 < 0 , divide by 2
5x^3 + 3x^2 - 45x - 27 < 0
x^2(5x+3) - 9(5x+3) < 0
(5x+3)(x^2-9) < 0
(5x+3)(x+3)(x-3) < 0
critical values are x = -5/3, -3, and 3
Test the expression for different values, one in each part of the domain.
We don't have to find the actual value, just the sign of the answer
let x = -5 , for x < -5/3
- - - < 0 , that works
let x = -2 , for between -5/3 and -3
- + - > 0 , no good
let x = 0 , for between -3 and 3
+ + - < 0 , that works
let x = 10, for x > 3
+++ > 0 , no good
so x < -5/3 OR -3 < x < 3
I will let you change that to the interval notation that you were taught. Personally I prefer the above notation.
Me too but it gets counted wrong in this format.