Solve the inequality (x - 1)(x + 2)(x - 3) < 0 and write the solution set in interval notation.

Step-by-step solution

1. 1

   Do not expand the product. The inequality is already factored, and that is exactly the form you want. Multiplying it out gives $x^3-2x^2-5x+6$, which tells you nothing about signs. A product of factors is negative or positive according to how many of its factors are negative, so keep the factored form and study one factor at a time.

2. 2

   Find the zeros of every factor. Setting each factor to zero gives

   $x-1=0\Rightarrow x=1,\qquad x+2=0\Rightarrow x=-2,\qquad x-3=0\Rightarrow x=3$

   These are the only places where the product can change sign, because a continuous function can only switch from negative to positive by passing through $0$.

3. 3

   Split the number line at $-2$, $1$ and $3$. The three zeros cut the real line into four open intervals:

   $(-\infty,-2),\quad (-2,1),\quad (1,3),\quad (3,+\infty)$

   Inside each interval the sign of the product is constant, so a single test point decides the whole interval.

4. 4

   Test one point per interval. Record the sign of each factor rather than the numerical value:

   interval	test $x$	$x+2$	$x-1$	$x-3$	product
   $(-\infty,-2)$	$-3$	$-$	$-$	$-$	$-$
   $(-2,1)$	$0$	$+$	$-$	$-$	$+$
   $(1,3)$	$2$	$+$	$+$	$-$	$-$
   $(3,+\infty)$	$4$	$+$	$+$	$+$	$+$

   Three negatives multiply to a negative, two negatives to a positive: that is the whole rule.

5. 5

   Notice the alternating pattern. Because each factor appears to the first power (every root is simple), the sign flips at every zero. Starting from the far right, where all factors are positive, the signs run $+,-,+,-$ as you move left. Once you know the sign of the rightmost interval you can fill in the rest without testing anything — the table above is just a check.

6. 6

   Collect the intervals where the product is negative, and exclude the endpoints. The product is negative on $(-\infty,-2)$ and on $(1,3)$. The inequality is strict, so $x=-2$, $x=1$ and $x=3$ (where the product equals $0$, not less than $0$) are left out:

   $\boxed{(-\infty,-2)\cup(1,3)}$

   Spot check: $x=-3$ gives $(-4)(-1)(-6)=-24<0$ and $x=2$ gives $(1)(4)(-1)=-4<0$, while $x=0$ gives $(-1)(2)(-3)=6>0$, as predicted.