Answer
Solution set: $ [-2,\displaystyle \frac{1}{3}]$
Work Step by Step
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}]$
