Precalculus Examples

$\frac{2 z^{3} + 6 z^{2} + 18 z}{z^{3} - 27}$

Step 1

Factor $2 z$ out of $2 z^{3} + 6 z^{2} + 18 z$ .

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Step 1.1

Factor $2 z$ out of $2 z^{3}$ .

$\frac{2 z (z^{2}) + 6 z^{2} + 18 z}{z^{3} - 27}$

Step 1.2

Factor $2 z$ out of $6 z^{2}$ .

$\frac{2 z (z^{2}) + 2 z (3 z) + 18 z}{z^{3} - 27}$

Step 1.3

Factor $2 z$ out of $18 z$ .

$\frac{2 z (z^{2}) + 2 z (3 z) + 2 z (9)}{z^{3} - 27}$

Step 1.4

Factor $2 z$ out of $2 z (z^{2}) + 2 z (3 z)$ .

$\frac{2 z (z^{2} + 3 z) + 2 z (9)}{z^{3} - 27}$

Step 1.5

Factor $2 z$ out of $2 z (z^{2} + 3 z) + 2 z (9)$ .

$\frac{2 z (z^{2} + 3 z + 9)}{z^{3} - 27}$

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $27$ as $3^{3}$ .

$\frac{2 z (z^{2} + 3 z + 9)}{z^{3} - 3^{3}}$

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 $a = z$ and $b = 3$ .

$\frac{2 z (z^{2} + 3 z + 9)}{(z - 3) (z^{2} + z \cdot 3 + 3^{2})}$

Step 2.3

Simplify.

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Step 2.3.1

Move $3$ to the left of $z$ .

$\frac{2 z (z^{2} + 3 z + 9)}{(z - 3) (z^{2} + 3 \cdot z + 3^{2})}$

Step 2.3.2

Raise $3$ to the power of $2$ .

$\frac{2 z (z^{2} + 3 z + 9)}{(z - 3) (z^{2} + 3 z + 9)}$

Step 3

Cancel the common factor of $z^{2} + 3 z + 9$ .

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Step 3.1

Cancel the common factor.

$\frac{2 z (z^{2} + 3 z + 9)}{(z - 3) (z^{2} + 3 z + 9)}$

Step 3.2

Rewrite the expression.

$\frac{2 z}{z - 3}$