Precalculus Examples

$\frac{x^{3} - 64}{x^{2} - 3 x - 4}$

Step 1

Simplify the numerator.

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Step 1.1

Rewrite $64$ as $4^{3}$ .

$\frac{x^{3} - 4^{3}}{x^{2} - 3 x - 4}$

Step 1.2

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 $b = 4$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Move $4$ to the left of $x$ .

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

Step 1.3.2

Raise $4$ to the power of $2$ .

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

Step 2

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

$- 4, 1$

Step 2.2

Write the factored form using these integers.

$\frac{(x - 4) (x^{2} + 4 x + 16)}{(x - 4) (x + 1)}$

Step 3

Cancel the common factor of $x - 4$ .

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Step 3.1

Cancel the common factor.

$\frac{(x - 4) (x^{2} + 4 x + 16)}{(x - 4) (x + 1)}$

Step 3.2

Rewrite the expression.

$\frac{x^{2} + 4 x + 16}{x + 1}$