Precalculus Examples

$\frac{z^{3} - 8}{z^{2} + 2 z + 4}$

Step 1

Simplify the numerator.

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Step 1.1

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

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

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 $a = z$ and $b = 2$ .

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

Step 1.3

Simplify.

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 2

Cancel the common factor of $z^{2} + 2 z + 4$ .

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Step 2.1

Cancel the common factor.

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

Step 2.2

Divide $z - 2$ by $1$ .

$z - 2$