Proof by Contradiction (I think) Prove: If $x^2+x-6 \ge 0$, then  $x\le -3$ or $x \ge 2.$
This is my initial work, but I'm not sure if this is the way to go or if should do a direct proof.
Proof by contradiction: Assume $x^2+x-6 \ge 0$ and $-3&lt;x&lt;2$.  Is this the correct start?
$f(x)=x^2+x-6=(x+3)(x-2)$
From that one can see that $f(x)\geq 0$ when $x\geq 2$ of $x\leq -3$.
As you have assumed, let $-3&lt;x&lt;2$ which implies that $x+3&gt;0$ and $x-2&lt;0$, i.e. these are quantities of opposite signs, so their product must be negative. But, $(x+3)(x-2)=x^2+x-6$ is non-negative, as given, hence, contradiction!
An algorithmic way of looking at questions of this type is given the name wavy-curve method.
In the direct way, $x^2+x-6\ge 0 \implies (x+3)(x-2)\ge 0$, so both the factors must be of the same sign.
There are only really $3$ regions to check for signs. $(-\infty,-3],(-3,2),[2,\infty)$.
In $[2,\infty)$, both $x+3$ and $x-2$ are non-negative, so their product is also non-negative. As you go into $(-3,2)$, the sign of $x+3$ remains the same, i.e. it is still non-negative, but the sign of $x-2$ is now negative, so their product is now negative. Going in $(-\infty,-3)$, the sign of $x-2$ is still negative, as it was in the previous interval, but that of $x+3$ has changed to negative as well, so their product is now positive.
Thus, the given term $x^2+x-6$ changes signs at the junctions of crossover $(2,3)$, and it is non-negative in two of them.
You have to be careful with choosing the endpoints of the intervals, which depends on whether there is an equality sign or not in the given inequality.
Also, try this: For what $x$ is $(x-1)(x-2)^2(x-3)^3&gt;0$? (specify regions, that is ...)