Find the Holes in the Graph (x^2-7x+12)/(-5x+15)

$\frac{x^{2} - 7 x + 12}{- 5 x + 15}$

Step 1

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.

$\frac{(x - 4) (x - 3)}{- 5 x + 15}$

Step 2

Factor $5$ out of $- 5 x + 15$ .

Factor $5$ out of $- 5 x$ .

$\frac{(x - 4) (x - 3)}{5 (- x) + 15}$

Factor $5$ out of $15$ .

$\frac{(x - 4) (x - 3)}{5 (- x) + 5 (3)}$

Factor $5$ out of $5 (- x) + 5 (3)$ .

$\frac{(x - 4) (x - 3)}{5 (- x + 3)}$

Step 3

Cancel the common factor of $x - 3$ and $- x + 3$ .

Factor $- 1$ out of $x$ .

$\frac{(x - 4) (- 1 (- x) - 3)}{5 (- x + 3)}$

Rewrite $- 3$ as $- 1 (3)$ .

$\frac{(x - 4) (- 1 (- x) - 1 (3))}{5 (- x + 3)}$

Factor $- 1$ out of $- 1 (- x) - 1 (3)$ .

$\frac{(x - 4) (- 1 (- x + 3))}{5 (- x + 3)}$

Cancel the common factor.

$\frac{(x - 4) (- 1 (- x + 3))}{5 (- x + 3)}$

Rewrite the expression.

$\frac{(x - 4) \cdot (- 1)}{5}$

Step 4

Multiply $x - 4$ by $- 1$ .

$\frac{(x - 4) \cdot - 1}{5}$

Step 5

Move $- 1$ to the left of $x - 4$ .

$\frac{- 1 \cdot (x - 4)}{5}$

Step 6

Move the negative in front of the fraction.

$- \frac{x - 4}{5}$

Step 7

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

$- x + 3$

Step 8

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{x - 4}{5}$ .

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

$- x + 3 = 0$

Solve $- x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

Substitute $3$ for $x$ in $- \frac{x - 4}{5}$ and simplify.

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

$- \frac{3 - 4}{5}$

Simplify.

Subtract $4$ from $3$ .

$- \frac{- 1}{5}$

Move the negative in front of the fraction.

$- - \frac{1}{5}$

Multiply $- - \frac{1}{5}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{1}{5})$

Multiply $\frac{1}{5}$ by $1$ .

$\frac{1}{5}$

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

$(3, \frac{1}{5})$

Step 9

Find the Holes in the Graph (x^2-7x+12)/(-5x+15)

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