Precalculus Examples

$\frac{x^{2} - x - 12}{16 - x^{2}}$

Step 1

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Step 1.2

Write the factored form using these integers.

$\frac{(x - 4) (x + 3)}{16 - x^{2}}$

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $16$ as $4^{2}$ .

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

Step 2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = x$ .

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

Step 3

Reduce the expression by cancelling the common factors.

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Step 3.1

Cancel the common factor of $x - 4$ and $4 - x$ .

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Step 3.1.1

Factor $- 1$ out of $x$ .

$\frac{(- 1 (- x) - 4) (x + 3)}{(4 + x) (4 - x)}$

Step 3.1.2

Rewrite $- 4$ as $- 1 (4)$ .

$\frac{(- 1 (- x) - 1 (4)) (x + 3)}{(4 + x) (4 - x)}$

Step 3.1.3

Factor $- 1$ out of $- 1 (- x) - 1 (4)$ .

$\frac{- 1 (- x + 4) (x + 3)}{(4 + x) (4 - x)}$

Step 3.1.4

Reorder terms.

$\frac{- 1 (- x + 4) (x + 3)}{(4 + x) (- x + 4)}$

Step 3.1.5

Cancel the common factor.

$\frac{- 1 (- x + 4) (x + 3)}{(4 + x) (- x + 4)}$

Step 3.1.6

Rewrite the expression.

$\frac{- 1 (x + 3)}{4 + x}$

Step 3.2

Move the negative in front of the fraction.

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