The critical values are

`x = −3`, `x = 2` and `x = 4`.

These critical values divide the number line into 4 intervals.

Hence, the signs of the function are:

Interval | `((x−2)^2(x+3))/(4−x)` | sign of
f(x) |

`x < −3` | − | |

`−3 < x < 2` | + | |

`2 < x < 4` | + | |

`x > 4` | − |

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

Hence, the solution is: `x < −3` or `x > 4`.

Here's the graph of the solution.

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