Find the Holes in the Graph (-4x+8)/(x^2-7x+10)

$\frac{- 4 x + 8}{x^{2} - 7 x + 10}$

Step 1

Factor $4$ out of $- 4 x + 8$ .

Factor $4$ out of $- 4 x$ .

$\frac{4 (- x) + 8}{x^{2} - 7 x + 10}$

Factor $4$ out of $8$ .

$\frac{4 (- x) + 4 (2)}{x^{2} - 7 x + 10}$

Factor $4$ out of $4 (- x) + 4 (2)$ .

$\frac{4 (- x + 2)}{x^{2} - 7 x + 10}$

Step 2

Factor $x^{2} - 7 x + 10$ 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 $10$ and whose sum is $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

$\frac{4 (- x + 2)}{(x - 5) (x - 2)}$

Step 3

Cancel the common factor of $- x + 2$ and $x - 2$ .

Factor $- 1$ out of $- x$ .

$\frac{4 (- (x) + 2)}{(x - 5) (x - 2)}$

Rewrite $2$ as $- 1 (- 2)$ .

$\frac{4 (- (x) - 1 (- 2))}{(x - 5) (x - 2)}$

Factor $- 1$ out of $- (x) - 1 (- 2)$ .

$\frac{4 (- (x - 2))}{(x - 5) (x - 2)}$

Rewrite $- (x - 2)$ as $- 1 (x - 2)$ .

$\frac{4 (- 1 (x - 2))}{(x - 5) (x - 2)}$

Cancel the common factor.

$\frac{4 (- 1 (x - 2))}{(x - 5) (x - 2)}$

Rewrite the expression.

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

Step 4

Simplify the numerator.

Multiply $4$ by $- 1$ .

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

Multiply $4$ by $- 1$ .

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

Step 5

Move the negative in front of the fraction.

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

Step 6

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

$x - 2$

Step 7

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

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $- \frac{4}{x - 5}$ and simplify.

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

$- \frac{4}{2 - 5}$

Simplify.

Subtract $5$ from $2$ .

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{4}{3}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{4}{3})$

Multiply $\frac{4}{3}$ by $1$ .

$\frac{4}{3}$

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

$(2, \frac{4}{3})$

Step 8

Find the Holes in the Graph (-4x+8)/(x^2-7x+10)

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