Find the Holes in the Graph (x-1)/(2x^2-22x+20)

$\frac{x - 1}{2 x^{2} - 22 x + 20}$

Step 1

Factor $2 x^{2} - 22 x + 20$ .

Factor $2$ out of $2 x^{2} - 22 x + 20$ .

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

$\frac{x - 1}{2 (x^{2}) - 22 x + 20}$

Factor $2$ out of $- 22 x$ .

$\frac{x - 1}{2 (x^{2}) + 2 (- 11 x) + 20}$

Factor $2$ out of $20$ .

$\frac{x - 1}{2 x^{2} + 2 (- 11 x) + 2 \cdot 10}$

Factor $2$ out of $2 x^{2} + 2 (- 11 x)$ .

$\frac{x - 1}{2 (x^{2} - 11 x) + 2 \cdot 10}$

Factor $2$ out of $2 (x^{2} - 11 x) + 2 \cdot 10$ .

$\frac{x - 1}{2 (x^{2} - 11 x + 10)}$

Factor.

Factor $x^{2} - 11 x + 10$ 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 $- 11$ .

$- 10, - 1$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 2

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

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

$x - 1$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{1}{2 (x - 10)}$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

Substitute $1$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Subtract $10$ from $1$ .

$\frac{1}{2 \cdot - 9}$

Multiply $2$ by $- 9$ .

$\frac{1}{- 18}$

Move the negative in front of the fraction.

$- \frac{1}{18}$

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

$(1, - \frac{1}{18})$

Step 5

Find the Holes in the Graph (x-1)/(2x^2-22x+20)

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