Precalculus Examples

$\frac{x^{2} + 6 x - 16}{x^{3} - 8}$

Step 1

Factor $x^{2} + 6 x - 16$ 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 $- 16$ and whose sum is $6$ .

$- 2, 8$

Step 1.2

Write the factored form using these integers.

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

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $8$ as $2^{3}$ .

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

Step 2.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 2$ .

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

Step 2.3

Simplify.

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Step 2.3.1

Move $2$ to the left of $x$ .

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

Step 2.3.2

Raise $2$ to the power of $2$ .

$\frac{(x - 2) (x + 8)}{(x - 2) (x^{2} + 2 x + 4)}$

Step 3

Cancel the common factor of $x - 2$ .

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Step 3.1

Cancel the common factor.

$\frac{(x - 2) (x + 8)}{(x - 2) (x^{2} + 2 x + 4)}$

Step 3.2

Rewrite the expression.

$\frac{x + 8}{x^{2} + 2 x + 4}$