(x+2)(x-4)(3+x)>0 interval notation

Question

im stuck on properly notating i know x is (-3,2) im confused by the middle

Answer 1

We need to find intervals where the function is positive.  This means that the function is above the axis.  Let us start by setting the function equal to zero, so that we can determine which x values make the function zero, or on the x-axis.

 

Set these factors of the function equal to zero.

 

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

x = -2                  x = 4                      x = -3

 

 

Our zeros are -3, -2, and 4.  Next, we will choose x values around these points and use them as test points to see where the function is positive or negative.

 

The test point we will use will be  x=-4 ,  x=-2.5 ,  x=0 , and x=5.

 

f(-4) = (-4 + 2)(-4 - 4)(3 + (-4))

        = (-2)(-8)(-1)

        = -16

 

f(-2.5) = (-2.5 + 2)(-2.5 - 4)(3 + (-2.5))

          = (-0.5)(-6.5)(0.5)

          = 1.625

 

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

      = (2)(-4)(3)

      = -24

 

f(5) = (5 + 2)(5 - 4)(3 + 5)

      = (7)(1)(8)

      = 56

 

Based on the test points, the function is positive between x=-3 and x=-2.  It is also positive when x is greater than 4.

 

In interval notation, the solution is

 

(-3, -2)∪(4, ∞).

Answer 2

To make the product of these three factors positive, simply make x-4 positive. So, x must be greater than 4.

 

(4, infinity) as an interval