To find the domain of the expression:

$\frac{(x+3)(x+1)}{(x+1)(x-2)}$

we need to identify the values of $x$ that would make the denominator zero, as these values are not allowed in the domain.

The denominator is:

$(x+1)(x-2)$

Setting each factor in the denominator equal to zero:

1. $x + 1 = 0 \Rightarrow x = -1$
2. $x - 2 = 0 \Rightarrow x = 2$

Thus, $x = -1$ and $x = 2$ are values that make the denominator zero, which means they are excluded from the domain.

Domain: All real numbers $x$ except $x = -1$ and $x = 2$.

In interval notation, the domain is:

$(-\infty, -1) \cup (-1, 2) \cup (2, \infty)$

Would you like more details, or do you have any questions?

Here are some related questions to expand on this topic:

1. How would the domain change if the expression in the numerator were different?
2. What is the range of this rational function?
3. How does simplifying the expression affect the domain?
4. How can you find the vertical asymptotes of this function?
5. What is the significance of removable discontinuities in rational functions?

Tip: When analyzing rational functions, always start by identifying points that make the denominator zero, as they determine restrictions on the domain.