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March 30, 2015

March 30, 2015

Posted by **anon** on Saturday, December 10, 2011 at 10:27pm.

10^3 + 6x^2 - 90x - 54 < 0

- algebra -
**Reiny**, Saturday, December 10, 2011 at 10:56pmI 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.

- algebra -
**anon**, Sunday, December 11, 2011 at 1:06pmMe too but it gets counted wrong in this format.

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