Using methods learnt in earlier chapters (see Remainder and Factor Theorems), the expression can be factored to give:

(x + 1)(x − 2)(x − 3) < 0

Setting the factors to zero, we get:

(x + 1) = 0

(x − 2) = 0

(x − 3) = 0

So `x = −1`, `x = 2`, or `x = 3`.

Therefore the critical values are

`x = −1`, `x = 2` and `x = 3`.

These critical values divide the number line into 4 intervals:

x < −1,

−1 < x < 2,

2 < x < 3 and

x > 3.

Next, we determine the sign of the function by the following method:

Interval `(x + 1)` `(x − 2)` `(x − 3)` sign of f(x)
`x < −1`
`−1 < x < 2` + +
`2 < x < 3` + +
`x > 3` + + + +

Since we want `f(x) < 0`, the intervals that satisfy this inequality will be those with a negative sign.

Hence, the solution is: `x < −1` or `2 < x < 3`.

Here is the graph of our solution:

123450-1-2-3-4xOpen image in a new page