Find the Holes in the Graph (x^2-4)/(x^2-5x+6)

Rewrite ${\displaystyle 4}$ as ${\displaystyle 2^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=x}$ and ${\displaystyle b=2}$.

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle 6}$ and whose sum is ${\displaystyle -5}$.

Write the factored form using these integers.

Cancel the common factor.

Rewrite the expression.

Set ${\displaystyle x-2}$ equal to ${\displaystyle 0}$.

Add ${\displaystyle 2}$ to both sides of the equation.

Substitute ${\displaystyle 2}$ for ${\displaystyle x}$ in ${\displaystyle \frac{x+2}{x-3}}$ and simplify.

The holes in the graph are the points where any of the cancelled factors are equal to ${\displaystyle 0}$.