I need to solve 3x^2+8x is less than or equal to 0 and then give the answer in interval notation, I cannot remember how to do this.

Question

 I just want to see it solved in steps so I can understand how to do the problem myself.

Answer 1

We set an inequality.

 

 

3x² + 8x ≤ 0

 

 

Now we factor.

 

x(3x + 8) ≤ 0

 

x ≤ 0      or    3x + 8 ≤ 0

                     3x ≤ -8

 

                      x ≤ -8/3

 

We perform a test point to see which value of x from our solutions will make the polynomial less than or equal to zero.

 

We will test x = -1, x = 1, x = -9/3, and x = -7/3

 

f(-1) = -1[3(-1) + 8]

        = -1(5)

        = -5

 

f(1) = 1[3(1) + 8]

      = 1(11)

      = 11

 

f(-9/3) = (-9/3)[3(-9/3) + 8]

           = (-3)(-9+ 8)

           = (-3)(-1)

           = 3

 

f(-7/3) = (-7/3)[3(-7/3) + 8]

          = (-7/3)(-7 + 8)

          = (-7/3)(1)

          = -7/3

 

Based on these test points, we accept x ≤ 0 and x ≥ -8/3.

 

In interval notation  [-8/3, 0].