Question

solve for x and graph the solution.
3(x + 2)(x - 3) < 0
plot the endpoints. select an endpoint to change it from closed to open. select the middle of a segment, ray, or line to delete it.

Explanation:

Step1: Find the critical points

Set \(3(x + 2)(x - 3)=0\). Using the zero - product property \(a\times b\times c = 0\) implies \(a = 0\) or \(b = 0\) or \(c = 0\). Here, \(x+2 = 0\) gives \(x=-2\) and \(x - 3=0\) gives \(x = 3\). These are the critical points that divide the number line into three intervals: \((-\infty,-2)\), \((-2,3)\), and \((3,\infty)\).

Step2: Test intervals

• For the interval \((-\infty,-2)\), let \(x=-3\). Then \(3(-3 + 2)(-3 - 3)=3\times(-1)\times(-6)=18>0\).
• For the interval \((-2,3)\), let \(x = 0\). Then \(3(0 + 2)(0 - 3)=3\times2\times(-3)=-18<0\).
• For the interval \((3,\infty)\), let \(x = 4\). Then \(3(4 + 2)(4 - 3)=3\times6\times1 = 18>0\).

Since the inequality is \(3(x + 2)(x - 3)<0\), the solution of the inequality is the interval where the expression is negative.

Answer:

The solution of the inequality \(3(x + 2)(x - 3)<0\) is \(x\in(-2,3)\). On the number - line, we plot open circles at \(x=-2\) and \(x = 3\) (because the inequality is strict, i.e., \(<\) not \(\leq\)) and shade the region between them.