Pre-Algebra Examples

$\frac{\frac{{(x - 8)}^{7} (x + 5)}{x^{2} + 6 x + 5}}{\frac{x^{2} - 16 x + 64}{x^{2} + 10 x + 25}}$

Step 1

Multiply the numerator by the reciprocal of the denominator.

$\frac{{(x - 8)}^{7} (x + 5)}{x^{2} + 6 x + 5} \cdot \frac{x^{2} + 10 x + 25}{x^{2} - 16 x + 64}$

Step 2

Factor $x^{2} + 6 x + 5$ 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 $5$ and whose sum is $6$ .

$1, 5$

Step 2.2

Write the factored form using these integers.

$\frac{{(x - 8)}^{7} (x + 5)}{(x + 1) (x + 5)} \cdot \frac{x^{2} + 10 x + 25}{x^{2} - 16 x + 64}$

Step 3

Cancel the common factor of $x + 5$ .

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Step 3.1

Cancel the common factor.

$\frac{{(x - 8)}^{7} (x + 5)}{(x + 1) (x + 5)} \cdot \frac{x^{2} + 10 x + 25}{x^{2} - 16 x + 64}$

Step 3.2

Rewrite the expression.

$\frac{{(x - 8)}^{7}}{x + 1} \cdot \frac{x^{2} + 10 x + 25}{x^{2} - 16 x + 64}$

Step 4

Factor using the perfect square rule.

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Step 4.1

Rewrite $25$ as $5^{2}$ .

$\frac{{(x - 8)}^{7}}{x + 1} \cdot \frac{x^{2} + 10 x + 5^{2}}{x^{2} - 16 x + 64}$

Step 4.2

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

$10 x = 2 \cdot x \cdot 5$

Step 4.3

Rewrite the polynomial.

$\frac{{(x - 8)}^{7}}{x + 1} \cdot \frac{x^{2} + 2 \cdot x \cdot 5 + 5^{2}}{x^{2} - 16 x + 64}$

Step 4.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 5$ .

$\frac{{(x - 8)}^{7}}{x + 1} \cdot \frac{{(x + 5)}^{2}}{x^{2} - 16 x + 64}$

Step 5

Factor using the perfect square rule.

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Step 5.1

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

$\frac{{(x - 8)}^{7}}{x + 1} \cdot \frac{{(x + 5)}^{2}}{x^{2} - 16 x + 8^{2}}$

Step 5.2

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

$16 x = 2 \cdot x \cdot 8$

Step 5.3

Rewrite the polynomial.

$\frac{{(x - 8)}^{7}}{x + 1} \cdot \frac{{(x + 5)}^{2}}{x^{2} - 2 \cdot x \cdot 8 + 8^{2}}$

Step 5.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 8$ .

$\frac{{(x - 8)}^{7}}{x + 1} \cdot \frac{{(x + 5)}^{2}}{{(x - 8)}^{2}}$

Step 6

Cancel the common factor of ${(x - 8)}^{2}$ .

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Step 6.1

Factor ${(x - 8)}^{2}$ out of ${(x - 8)}^{7}$ .

$\frac{{(x - 8)}^{2} {(x - 8)}^{5}}{x + 1} \cdot \frac{{(x + 5)}^{2}}{{(x - 8)}^{2}}$

Step 6.2

Cancel the common factor.

$\frac{{(x - 8)}^{2} {(x - 8)}^{5}}{x + 1} \cdot \frac{{(x + 5)}^{2}}{{(x - 8)}^{2}}$

Step 6.3

Rewrite the expression.

$\frac{{(x - 8)}^{5}}{x + 1} {(x + 5)}^{2}$

Step 7

Combine $\frac{{(x - 8)}^{5}}{x + 1}$ and ${(x + 5)}^{2}$ .

$\frac{{(x - 8)}^{5} {(x + 5)}^{2}}{x + 1}$