Basic Math Examples

$\frac{z^{3} - 1}{2 z^{2} - z - 1} \div \frac{z^{2} + z + 1}{2 z^{2} + 9 z + 4}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{z^{3} - 1}{2 z^{2} - z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 2

Simplify the numerator.

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Step 2.1

Rewrite $1$ as $1^{3}$ .

$\frac{z^{3} - 1^{3}}{2 z^{2} - z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

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 $b = 1$ .

$\frac{(z - 1) (z^{2} + z \cdot 1 + 1^{2})}{2 z^{2} - z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 2.3

Simplify.

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Step 2.3.1

Multiply $z$ by $1$ .

$\frac{(z - 1) (z^{2} + z + 1^{2})}{2 z^{2} - z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 2.3.2

One to any power is one.

$\frac{(z - 1) (z^{2} + z + 1)}{2 z^{2} - z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 3

Factor by grouping.

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Step 3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

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Step 3.1.1

Factor $- 1$ out of $- z$ .

$\frac{(z - 1) (z^{2} + z + 1)}{2 z^{2} - (z) - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 3.1.2

Rewrite $- 1$ as $1$ plus $- 2$

$\frac{(z - 1) (z^{2} + z + 1)}{2 z^{2} + (1 - 2) z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 3.1.3

Apply the distributive property.

$\frac{(z - 1) (z^{2} + z + 1)}{2 z^{2} + 1 z - 2 z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 3.1.4

Multiply $z$ by $1$ .

$\frac{(z - 1) (z^{2} + z + 1)}{2 z^{2} + z - 2 z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$\frac{(z - 1) (z^{2} + z + 1)}{(2 z^{2} + z) - 2 z - 1} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 3.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{(z - 1) (z^{2} + z + 1)}{z (2 z + 1) - (2 z + 1)} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $2 z + 1$ .

$\frac{(z - 1) (z^{2} + z + 1)}{(2 z + 1) (z - 1)} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 4

Cancel the common factor of $z^{2} + z + 1$ .

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Step 4.1

Factor $z^{2} + z + 1$ out of $(z - 1) (z^{2} + z + 1)$ .

$\frac{(z^{2} + z + 1) (z - 1)}{(2 z + 1) (z - 1)} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 4.2

Cancel the common factor.

$\frac{(z^{2} + z + 1) (z - 1)}{(2 z + 1) (z - 1)} \cdot \frac{2 z^{2} + 9 z + 4}{z^{2} + z + 1}$

Step 4.3

Rewrite the expression.

$\frac{z - 1}{(2 z + 1) (z - 1)} (2 z^{2} + 9 z + 4)$

Step 5

Cancel the common factor of $z - 1$ .

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Step 5.1

Cancel the common factor.

$\frac{z - 1}{(2 z + 1) (z - 1)} (2 z^{2} + 9 z + 4)$

Step 5.2

Rewrite the expression.

$\frac{1}{2 z + 1} (2 z^{2} + 9 z + 4)$

Step 6

Multiply $\frac{1}{2 z + 1}$ by $2 z^{2} + 9 z + 4$ .

$\frac{2 z^{2} + 9 z + 4}{2 z + 1}$

Step 7

Factor by grouping.

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Step 7.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 4 = 8$ and whose sum is $b = 9$ .

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Step 7.1.1

Factor $9$ out of $9 z$ .

$\frac{2 z^{2} + 9 (z) + 4}{2 z + 1}$

Step 7.1.2

Rewrite $9$ as $1$ plus $8$

$\frac{2 z^{2} + (1 + 8) z + 4}{2 z + 1}$

Step 7.1.3

Apply the distributive property.

$\frac{2 z^{2} + 1 z + 8 z + 4}{2 z + 1}$

Step 7.1.4

Multiply $z$ by $1$ .

$\frac{2 z^{2} + z + 8 z + 4}{2 z + 1}$

Step 7.2

Factor out the greatest common factor from each group.

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Step 7.2.1

Group the first two terms and the last two terms.

$\frac{(2 z^{2} + z) + 8 z + 4}{2 z + 1}$

Step 7.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{z (2 z + 1) + 4 (2 z + 1)}{2 z + 1}$

Step 7.3

Factor the polynomial by factoring out the greatest common factor, $2 z + 1$ .

$\frac{(2 z + 1) (z + 4)}{2 z + 1}$

Step 8

Cancel the common factor of $2 z + 1$ .

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Step 8.1

Cancel the common factor.

$\frac{(2 z + 1) (z + 4)}{2 z + 1}$

Step 8.2

Divide $z + 4$ by $1$ .

$z + 4$