Basic Math Examples

$\frac{z^{2} - 64}{z^{2} - 14 z + 49} \div \frac{6 z - 48}{z^{2} - 6 z - 7}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{z^{2} - 64}{z^{2} - 14 z + 49} \cdot \frac{z^{2} - 6 z - 7}{6 z - 48}$

Step 2

Simplify the numerator.

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Step 2.1

Rewrite $64$ as $8^{2}$ .

$\frac{z^{2} - 8^{2}}{z^{2} - 14 z + 49} \cdot \frac{z^{2} - 6 z - 7}{6 z - 48}$

Step 2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = z$ and $b = 8$ .

$\frac{(z + 8) (z - 8)}{z^{2} - 14 z + 49} \cdot \frac{z^{2} - 6 z - 7}{6 z - 48}$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $49$ as $7^{2}$ .

$\frac{(z + 8) (z - 8)}{z^{2} - 14 z + 7^{2}} \cdot \frac{z^{2} - 6 z - 7}{6 z - 48}$

Step 3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$14 z = 2 \cdot z \cdot 7$

Step 3.3

Rewrite the polynomial.

$\frac{(z + 8) (z - 8)}{z^{2} - 2 \cdot z \cdot 7 + 7^{2}} \cdot \frac{z^{2} - 6 z - 7}{6 z - 48}$

Step 3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = z$ and $b = 7$ .

$\frac{(z + 8) (z - 8)}{{(z - 7)}^{2}} \cdot \frac{z^{2} - 6 z - 7}{6 z - 48}$

Step 4

Factor $z^{2} - 6 z - 7$ using the AC method.

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Step 4.1

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 $- 7$ and whose sum is $- 6$ .

$- 7, 1$

Step 4.2

Write the factored form using these integers.

$\frac{(z + 8) (z - 8)}{{(z - 7)}^{2}} \cdot \frac{(z - 7) (z + 1)}{6 z - 48}$

Step 5

Simplify terms.

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Step 5.1

Factor $6$ out of $6 z - 48$ .

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Step 5.1.1

Factor $6$ out of $6 z$ .

$\frac{(z + 8) (z - 8)}{{(z - 7)}^{2}} \cdot \frac{(z - 7) (z + 1)}{6 (z) - 48}$

Step 5.1.2

Factor $6$ out of $- 48$ .

$\frac{(z + 8) (z - 8)}{{(z - 7)}^{2}} \cdot \frac{(z - 7) (z + 1)}{6 z + 6 \cdot - 8}$

Step 5.1.3

Factor $6$ out of $6 z + 6 \cdot - 8$ .

$\frac{(z + 8) (z - 8)}{{(z - 7)}^{2}} \cdot \frac{(z - 7) (z + 1)}{6 (z - 8)}$

Step 5.2

Combine.

$\frac{(z + 8) (z - 8) ((z - 7) (z + 1))}{{(z - 7)}^{2} (6 (z - 8))}$

Step 5.3

Cancel the common factor of $z - 8$ .

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Step 5.3.1

Cancel the common factor.

$\frac{(z + 8) (z - 8) ((z - 7) (z + 1))}{{(z - 7)}^{2} (6 (z - 8))}$

Step 5.3.2

Rewrite the expression.

$\frac{(z + 8) ((z - 7) (z + 1))}{{(z - 7)}^{2} \cdot (6)}$

Step 5.4

Cancel the common factor of $z - 7$ and ${(z - 7)}^{2}$ .

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Step 5.4.1

Factor $z - 7$ out of $(z + 8) ((z - 7) (z + 1))$ .

$\frac{(z - 7) ((z + 8) (z + 1))}{{(z - 7)}^{2} \cdot (6)}$

Step 5.4.2

Cancel the common factors.

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Step 5.4.2.1

Factor $z - 7$ out of ${(z - 7)}^{2} \cdot (6)$ .

$\frac{(z - 7) ((z + 8) (z + 1))}{(z - 7) ((z - 7) \cdot 6)}$

Step 5.4.2.2

Cancel the common factor.

$\frac{(z - 7) ((z + 8) (z + 1))}{(z - 7) ((z - 7) \cdot 6)}$

Step 5.4.2.3

Rewrite the expression.

$\frac{(z + 8) (z + 1)}{(z - 7) \cdot 6}$

Step 5.5

Move $6$ to the left of $z - 7$ .

$\frac{(z + 8) (z + 1)}{6 (z - 7)}$