Find the Holes in the Graph (5x-20)/(x^2-9x+20)

$\frac{5 x - 20}{x^{2} - 9 x + 20}$

Step 1

Factor $5$ out of $5 x - 20$ .

Factor $5$ out of $5 x$ .

$\frac{5 (x) - 20}{x^{2} - 9 x + 20}$

Factor $5$ out of $- 20$ .

$\frac{5 x + 5 \cdot - 4}{x^{2} - 9 x + 20}$

Factor $5$ out of $5 x + 5 \cdot - 4$ .

$\frac{5 (x - 4)}{x^{2} - 9 x + 20}$

Step 2

Factor $x^{2} - 9 x + 20$ 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 $20$ and whose sum is $- 9$ .

$- 5, - 4$

Write the factored form using these integers.

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

Step 3

Cancel the common factor of $x - 4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x - 4$

Step 5

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

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

Substitute $4$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Subtract $5$ from $4$ .

$\frac{5}{- 1}$

Divide $5$ by $- 1$ .

$- 5$

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

$(4, - 5)$

Step 6

Find the Holes in the Graph (5x-20)/(x^2-9x+20)

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