## College Algebra (6th Edition)

Solution set: $(-\displaystyle \infty, -\frac{3}{2})\cup(0,\infty)$ Follow the "Procedure for Solving Polynomial lnequalities",\ p.412: 1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$ where $f$ is a polynomial function. $2x^{2}+3x > 0$ $f(x)=2x^{2}+3x \quad$...factor the trinomial... $f(x)=x(2x+3)$ $x(2x+3) > 0$ 2. Solve the equation $f(x)=0$. The real solutions are the boundary points. $x(2x+3) = 0$ $x=0$ or $x=-\displaystyle \frac{3}{2}$ 3. Locate these boundary points on a number line, thereby dividing the number line into intervals. 4. Test each interval's sign of $f(x)$ with a test value, $\begin{array}{llll} Intervals: & (-\infty, -\frac{3}{2}) & (-\frac{3}{2},0) & (0,\infty)\\ a=test.val. & -3 & -1 & 1\\ f(a) & (-3)(-3) & (-1)(1) & (1)(5)\\ f(a) > 0 ? & T & F & T \end{array}$ 5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points. Solution set: $(-\displaystyle \infty, -\frac{3}{2})\cup(0,\infty)$