Simplify the fraction with numerator x2 - 9 and denominator x + 3.

Step-by-step solution

1. 1

   Record the domain before simplifying. The denominator vanishes when $x+3=0$, so the expression is undefined at $x=-3$. Noting this first matters, because the simplification will hide it.

2. 2

   Factor the numerator as a difference of squares. Writing $9=3^2$ and applying $a^2-b^2=(a-b)(a+b)$:

   $x^2-9=(x-3)(x+3)$

   The factored form is what reveals the shared factor; no amount of term-by-term cancelling would find it.

3. 3

   Cancel the common factor.

   $\frac{(x-3)(x+3)}{x+3}=x-3$

   This is legitimate for every $x$ where the original is defined, i.e. wherever $x+3\neq 0$.

4. 4

   State the restriction with the answer.

   $\frac{x^2-9}{x+3}=x-3,\qquad x\neq -3$

   The two sides are not the same function: the left side has a hole at $x=-3$ while the line $x-3$ does not. This is the standard example of a removable discontinuity, and $\lim_{x\to-3}\frac{x^2-9}{x+3}=-6$ is the value the hole "wants".

5. 5

   Check with a value. At $x=1$: the original is $\frac{1-9}{4}=-2$, and $x-3=-2$ ✓. At $x=-3$ the original is $\frac00$, undefined, while $x-3$ would give $-6$ — confirming that the restriction is not optional.