Answer
Solution set: $[-4,\displaystyle \frac{2}{3}]$
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.
$f(x)=3x^{2}+10x-8\leq 0$
factor the trinomial...
find factors of $3(-8)=-24$ that add to $10:$
($12$ and $-2$ )
$3x^{2}+10x-8=3x^{2}+12x-2x-8= \quad$ ... in pairs ...
$=3x(x+4)-2(x+4)=(3x-2)(x+4)$
$f(x)=(3x-2)(x+4)\leq 0$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(3x-2)(x+4)=0$
$x=\displaystyle \frac{2}{3}$ or $x=-4$
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,
$\begin{array}{llll}
Intervals: & (-\infty, -4) & (-4,\frac{2}{3}) & (\frac{2}{3},\infty)\\
a=test.val. & -10 & 0 & 1\\
f(a) & (-32)(-6) & (-2)(4) & (1)(5)\\
f(a) \leq 0 ? & F & T & F
\end{array}$
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.
No intervals satisfy the inequality, border 3 is excluded...
Solution set: $[-4,\displaystyle \frac{2}{3}]$
