Question 135536
{{{x^3+2x^2-11x-12=0}}} Set the left side equal to zero. 



When we use a graphing calculator, we find that the solutions are {{{x=-4}}}, {{{x=-1}}} and {{{x=3}}}. So that means that the inequality {{{x^3+2x^2-11x-12<0}}} factors to {{{(x+4)(x+1)(x-3)<0}}}





Now plot the critical values on a number line


{{{number_line( 600, -10, 10, -4,-1,3 )}}}




So let's evaluate a point that is to the left of *[Tex \LARGE  x=-4 ] (which is the left most endpoint). Let's evaluate the value {{{x=-5}}}

  

*[Tex \LARGE (x+4)(x+1)(x-3)<0]  Start with the given inequality


*[Tex \LARGE (-5+4)(-5+1)(-5-3)<0]  Plug in {{{x=-5}}}


*[Tex \LARGE -32< 0]  Evaluate and simplify




Since *[Tex \LARGE -32< 0] is true, this means that one part of the solution in interval notation is *[Tex \LARGE \left(-\infty,-4\right)]

-----------------------------------------------



Now let's test a point that is in between the critical values *[Tex \LARGE  x=-4 ] and *[Tex \LARGE  x=-1 ]



*[Tex \LARGE (x+4)(x+1)(x-3)<0]  Start with the given inequality


*[Tex \LARGE (-2+4)(-2+1)(-2-3)<0]  Plug in {{{x=-2}}}


*[Tex \LARGE 10< 0]  Evaluate and simplify



 Since *[Tex \LARGE 10< 0] is false, this means that the interval does not work. So that means that this interval is not in the solution set and can be ignored.

-----------------------------------------------



Now let's test a point that is in between the critical values *[Tex \LARGE  x=-1 ] and *[Tex \LARGE  x=3 ]



*[Tex \LARGE (x+4)(x+1)(x-3)<0]  Start with the given inequality


*[Tex \LARGE (1+4)(1+1)(1-3)<0]  Plug in {{{x=1}}}


*[Tex \LARGE -20< 0]  Evaluate and simplify




Since *[Tex \LARGE -20< 0] is true, this means that one part of the solution in interval notation is *[Tex \LARGE \left(-1,3\right)]

-----------------------------------------------



So let's evaluate a point that is to the right of *[Tex \LARGE  x=3 ] (which is the right most endpoint). Let's evaluate the value {{{x=4}}}

  

*[Tex \LARGE (x+4)(x+1)(x-3)<0]  Start with the given inequality


*[Tex \LARGE (4+4)(4+1)(4-3)<0]  Plug in {{{x=4}}}


*[Tex \LARGE 40< 0]  Evaluate and simplify



 Since *[Tex \LARGE 40< 0] is false, this means that the interval does not work. So that means that this interval is not in the solution set and can be ignored.

-----------------------------------------------


Answer:

So the solution set in interval notation is *[Tex \LARGE \left(-\infty,-4\right)\cup\left(-1,3\right)]