## College Algebra 7th Edition

$(-\infty, 0] \cup [\frac{2}{3}, +\infty)$ Refer to the image below for the graph.
First step is to find the zeros of each factor. To find value of $x$ that will make each factor equal to zero, equate each factor to zero then solve each equation: $\begin{array}{ccc} &x=0 &\text{ or } &2-3x=0 \\&x=0 &\text{ or } &2=3x \\&x=0 &\text{ or } &\frac{2}{3}=x \end{array}$ Next step is to find the intervals. The zeros $0$ and $\frac{2}{3}$ divide the number line into three intervals, namely: $(-\infty, 0), (0, \frac{2}{3}), \text{ and } (\frac{2}{3}, +\infty)$. $\bf\text{Make a table of signs}.$ (refer to the attached image below) $\bf\text{Solve}$ From the table of signs, it can be seen that $x(2-3x)\le 0$ in the intervals $(-\infty, 0)$ and $(\frac{2}{3}, +\infty)$. The inequality involves $\le$ therefore the endpoints $0$ and $\frac{2}{3}$ are part of the solution set. Thus, the solution set is $(-\infty, 0] \cup [\frac{2}{3}, +\infty)$. To graph this, plot solid dots at $0$ and $\frac{2}{3}$ then shade the region to the left of $0$ and the region to the right of $\frac{2}{3}$. (refer to the attached image in the answer part above)