Find the Holes in the Graph (x+10)/(2x^2+38x+180)

$\frac{x + 10}{2 x^{2} + 38 x + 180}$

Step 1

Factor $2 x^{2} + 38 x + 180$ .

Factor $2$ out of $2 x^{2} + 38 x + 180$ .

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

$\frac{x + 10}{2 (x^{2}) + 38 x + 180}$

Factor $2$ out of $38 x$ .

$\frac{x + 10}{2 (x^{2}) + 2 (19 x) + 180}$

Factor $2$ out of $180$ .

$\frac{x + 10}{2 x^{2} + 2 (19 x) + 2 \cdot 90}$

Factor $2$ out of $2 x^{2} + 2 (19 x)$ .

$\frac{x + 10}{2 (x^{2} + 19 x) + 2 \cdot 90}$

Factor $2$ out of $2 (x^{2} + 19 x) + 2 \cdot 90$ .

$\frac{x + 10}{2 (x^{2} + 19 x + 90)}$

Factor.

Factor $x^{2} + 19 x + 90$ 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 $90$ and whose sum is $19$ .

$9, 10$

Write the factored form using these integers.

$\frac{x + 10}{2 ((x + 9) (x + 10))}$

Remove unnecessary parentheses.

$\frac{x + 10}{2 (x + 9) (x + 10)}$

Step 2

Cancel the common factor of $x + 10$ .

Cancel the common factor.

$\frac{x + 10}{2 (x + 9) (x + 10)}$

Rewrite the expression.

$\frac{1}{2 (x + 9)}$

Step 3

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

$x + 10$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{2 (x + 9)}$ .

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

Substitute $- 10$ for $x$ in $\frac{1}{2 (x + 9)}$ and simplify.

Substitute $- 10$ for $x$ to find the $y$ coordinate of the hole.

$\frac{1}{2 (- 10 + 9)}$

Simplify.

Add $- 10$ and $9$ .

$\frac{1}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$\frac{1}{- 2}$

Move the negative in front of the fraction.

$- \frac{1}{2}$

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

$(- 10, - \frac{1}{2})$

Step 5

Find the Holes in the Graph (x+10)/(2x^2+38x+180)

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