Find the Holes in the Graph (x^2-9x+20)/(4x^2-12x-40)

$\frac{x^{2} - 9 x + 20}{4 x^{2} - 12 x - 40}$

Step 1

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.

$\frac{(x - 5) (x - 4)}{4 x^{2} - 12 x - 40}$

Step 2

Factor $4 x^{2} - 12 x - 40$ .

Factor $4$ out of $4 x^{2} - 12 x - 40$ .

Factor $4$ out of $4 x^{2}$ .

$\frac{(x - 5) (x - 4)}{4 (x^{2}) - 12 x - 40}$

Factor $4$ out of $- 12 x$ .

$\frac{(x - 5) (x - 4)}{4 (x^{2}) + 4 (- 3 x) - 40}$

Factor $4$ out of $- 40$ .

$\frac{(x - 5) (x - 4)}{4 x^{2} + 4 (- 3 x) + 4 \cdot - 10}$

Factor $4$ out of $4 x^{2} + 4 (- 3 x)$ .

$\frac{(x - 5) (x - 4)}{4 (x^{2} - 3 x) + 4 \cdot - 10}$

Factor $4$ out of $4 (x^{2} - 3 x) + 4 \cdot - 10$ .

$\frac{(x - 5) (x - 4)}{4 (x^{2} - 3 x - 10)}$

Factor.

Factor $x^{2} - 3 x - 10$ 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 $- 3$ .

$- 5, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

Cancel the common factor of $x - 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x - 5$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 4}{4 (x + 2)}$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

Substitute $5$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Subtract $4$ from $5$ .

$\frac{1}{4 (5 + 2)}$

Add $5$ and $2$ .

$\frac{1}{4 \cdot 7}$

Multiply $4$ by $7$ .

$\frac{1}{28}$

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

$(5, \frac{1}{28})$

Step 6

Find the Holes in the Graph (x^2-9x+20)/(4x^2-12x-40)

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