## College Algebra (6th Edition)

Solution set: $[-2,\displaystyle \frac{1}{3}]$
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. $5x \leq 2-3x^{2}$ $3x^{2}+5x-2 \leq 0$ $f(x)= 3x^{2}+5x-2 \quad$...factor the trinomial... find factors of $3(-2)=-6$ that add to $+5:$ ($6$ and $-1$ ) $3x^{2}+5x-2=3x^{2}+6x-x-2= \quad$ ... in pairs ... $=3x(x+2)+(x+2)=(3x+2)(x+2)$ $f(x)=(3x-1)(x+2) \leq 0$ 2. Solve the equation $f(x)=0$. The real solutions are the boundary points. $(3x-1)(x+2)=0$ $x=\displaystyle \frac{1}{3}$ or $x=-2$ 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, $\begin{array}{llll} Intervals: & (-\infty, -2) & (-2,\frac{1}{3}) & (\frac{1}{3},\infty)\\ a=test.val. & -10 & 0 & 1\\ f(a) & (-29)(-8) & (-1)(2) & (2)(3)\\ f(a) \leq 0 ? & F & T & F \end{array}$ 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: $[-2,\displaystyle \frac{1}{3}]$