Find the Holes in the Graph f(x)=(2x^2-2)/(x^2-4x+3)

$f (x) = \frac{2 x^{2} - 2}{x^{2} - 4 x + 3}$

Step 1

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

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

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

$f (x) = \frac{2 (x^{2}) - 2}{x^{2} - 4 x + 3}$

Factor $2$ out of $- 2$ .

$f (x) = \frac{2 (x^{2}) + 2 (- 1)}{x^{2} - 4 x + 3}$

Factor $2$ out of $2 (x^{2}) + 2 (- 1)$ .

$f (x) = \frac{2 (x^{2} - 1)}{x^{2} - 4 x + 3}$

Rewrite $1$ as $1^{2}$ .

$f (x) = \frac{2 (x^{2} - 1^{2})}{x^{2} - 4 x + 3}$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$f (x) = \frac{2 ((x + 1) (x - 1))}{x^{2} - 4 x + 3}$

Remove unnecessary parentheses.

$f (x) = \frac{2 (x + 1) (x - 1)}{x^{2} - 4 x + 3}$

Step 2

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$f (x) = \frac{2 (x + 1) (x - 1)}{(x - 3) (x - 1)}$

Step 3

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$f (x) = \frac{2 (x + 1) (x - 1)}{(x - 3) (x - 1)}$

Rewrite the expression.

$f (x) = \frac{2 (x + 1)}{x - 3}$

Step 4

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

$x - 1$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{2 (x + 1)}{x - 3}$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $\frac{2 (x + 1)}{x - 3}$ and simplify.

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

$\frac{2 (1 + 1)}{1 - 3}$

Simplify.

Add $1$ and $1$ .

$\frac{2 \cdot 2}{1 - 3}$

Subtract $3$ from $1$ .

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

Multiply $2$ by $2$ .

$\frac{4}{- 2}$

Divide $4$ by $- 2$ .

$- 2$

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

$(1, - 2)$

Step 6

Find the Holes in the Graph f(x)=(2x^2-2)/(x^2-4x+3)

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