Precalculus Examples

$\frac{x^{2} + x - 6}{x^{2} - 6 x + 8}$

Step 1

Factor $x^{2} + x - 6$ using the AC method.

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Step 1.1

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 1.2

Write the factored form using these integers.

$\frac{(x - 2) (x + 3)}{x^{2} - 6 x + 8}$

Step 2

Factor $x^{2} - 6 x + 8$ using the AC method.

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Step 2.1

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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Step 2.2

Write the factored form using these integers.

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

Step 3

Cancel the common factor of $x - 2$ .

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Step 3.1

Cancel the common factor.

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

Step 3.2

Rewrite the expression.

$\frac{x + 3}{x - 4}$

Step 4

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

$x - 2$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 3}{x - 4}$ .

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Step 5.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 5.3

Substitute $2$ for $x$ in $\frac{x + 3}{x - 4}$ and simplify.

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Step 5.3.1

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

$\frac{2 + 3}{2 - 4}$

Step 5.3.2

Simplify.

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Step 5.3.2.1

Add $2$ and $3$ .

$\frac{5}{2 - 4}$

Step 5.3.2.2

Subtract $4$ from $2$ .

$\frac{5}{- 2}$

Step 5.3.2.3

Move the negative in front of the fraction.

$- \frac{5}{2}$

Step 5.4

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

$(2, - \frac{5}{2})$

Step 6

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