Find the Holes in the Graph (x^2-7x+12)/(x^2-9)

$\frac{x^{2} - 7 x + 12}{x^{2} - 9}$

Factor $x^{2} - 7 x + 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$\frac{(x - 4) (x - 3)}{x^{2} - 9}$

Factor $x^{2} - 9$ .

Rewrite $9$ as $3^{2}$ .

$\frac{(x - 4) (x - 3)}{x^{2} - 3^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\frac{(x - 4) (x - 3)}{(x + 3) (x - 3)}$

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{(x - 4) (x - 3)}{(x + 3) (x - 3)}$

Rewrite the expression.

$\frac{x - 4}{x + 3}$

To find the holes in the graph, look at the denominator factors that were cancelled.

$x - 3$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 4}{x + 3}$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $\frac{x - 4}{x + 3}$ and simplify.

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3 - 4}{3 + 3}$

Simplify.

Subtract $4$ from $3$ .

$\frac{- 1}{3 + 3}$

Add $3$ and $3$ .

$\frac{- 1}{6}$

Move the negative in front of the fraction.

$- \frac{1}{6}$

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

$(3, - \frac{1}{6})$

Find the Holes in the Graph (x^2-7x+12)/(x^2-9)

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