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: