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\(4x^3 + 16x^2 - 9x - 36 >= 0\)Find a solution for 4x^3 + 16x^2 - 9x - 36>/= 0

Guest Mar 14, 2017
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4x^2 + 16x^2 - 9x - 36 ≥ 0   set this to 0

 

4x^2 + 16x^2 - 9x - 36 = 0

 

Factor

 

4x^2(x + 4) - 9(x + 4) = 0

 

(x + 4) (4x^2 - 9)   = 0

 

(x + 4)(2x + 3) (2x - 3) = 0

 

Set each factor to 0 and solve for x ....and x = -4, x = -3/2  and x = 3/2

 

Look at the graph, here : https://www.desmos.com/calculator/tiouq73sj4

 

Notice that the intervals that make the original inequality true  are [-4, -3/2 ] and [3/2, infinity )

 

 

cool cool cool

CPhill  Mar 14, 2017

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