To factorize the expression $\frac{x^2 - 9}{x^2 + 5x + 6}$, we’ll start by factoring both the numerator and the denominator separately.

Step 1: Factorize the Numerator

The numerator is $x^2 - 9$, which is a difference of squares: $x^2 - 9 = (x + 3)(x - 3)$

Step 2: Factorize the Denominator

The denominator is $x^2 + 5x + 6$, which is a simple quadratic expression. We need two numbers that multiply to $6$ and add up to $5$. These numbers are $2$ and $3$: $x^2 + 5x + 6 = (x + 2)(x + 3)$

Step 3: Rewrite the Expression

Now, substitute these factorizations back into the original expression: $\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{(x + 3)(x - 3)}{(x + 2)(x + 3)}$

Step 4: Simplify the Expression

We can cancel out $(x + 3)$ from the numerator and denominator, assuming $x \neq -3$ to avoid division by zero: $= \frac{x - 3}{x + 2}$

Final Answer

The simplified form is: $\frac{x - 3}{x + 2}$ with the condition that $x \neq -3$ and $x \neq -2$ to keep the expression defined.

Would you like further details on any step or have additional questions?

Here are some related questions you might explore:

1. How do you factorize the difference of squares in general?
2. Why does cancelling terms require checking for restrictions on $x$?
3. How can we verify that $x - 3$ over $x + 2$ is equivalent to the original expression?
4. What happens if we substitute specific values for $x$ in the original and simplified expressions?
5. How would you handle this factorization if it was written with addition instead of subtraction in the numerator?

Tip: Always check for values that make any denominator zero when simplifying algebraic fractions.