Answer
Solution set: $(-\infty, -3)\cup(2,\infty)$
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.
$f(x)= x^{2}+x-6>0$
factor the trinomial...
find factors of $-6$ that add to $1:$
$f(x)=(x+3)(x-2)>0$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(x+3)(x-2)=0$
$x=-3$ 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,
$\begin{array}{llll}
Intervals: & (-\infty, -3) & (-3,2) & (2,\infty)\\
a=test.val. & -10 & 0 & 10\\
f(a) & (-7)(-12) & (3)(-2) & (13)(8)\\
f(a) > 0 ? & T & F & T
\end{array}$
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: $(-\infty, -3)\cup(2,\infty)$