Basic Math Examples

$\frac{a^{2} - 16 a + 64}{a^{2} - 64} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 1

Factor using the perfect square rule.

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Step 1.1

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

$\frac{a^{2} - 16 a + 8^{2}}{a^{2} - 64} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 1.2

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

$16 a = 2 \cdot a \cdot 8$

Step 1.3

Rewrite the polynomial.

$\frac{a^{2} - 2 \cdot a \cdot 8 + 8^{2}}{a^{2} - 64} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 1.4

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

$\frac{{(a - 8)}^{2}}{a^{2} - 64} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 2

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

$\frac{{(a - 8)}^{2}}{a^{2} - 8^{2}} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 3

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

$\frac{{(a - 8)}^{2}}{(a + 8) (a - 8)} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 4

Reduce the expression $\frac{{(a - 8)}^{2}}{(a + 8) (a - 8)}$ by cancelling the common factors.

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Step 4.1

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

$\frac{(a - 8) (a - 8)}{(a + 8) (a - 8)} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 4.2

Factor $a - 8$ out of $(a + 8) (a - 8)$ .

$\frac{(a - 8) (a - 8)}{(a - 8) (a + 8)} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 4.3

Cancel the common factor.

$\frac{(a - 8) (a - 8)}{(a - 8) (a + 8)} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 4.4

Rewrite the expression.

$\frac{a - 8}{a + 8} \cdot \frac{a^{3} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 5

Factor $a$ out of $a^{3} - 9 a^{2} + 8 a$ .

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Step 5.1

Factor $a$ out of $a^{3}$ .

$\frac{a - 8}{a + 8} \cdot \frac{a \cdot a^{2} - 9 a^{2} + 8 a}{2 a^{2} - 128}$

Step 5.2

Factor $a$ out of $- 9 a^{2}$ .

$\frac{a - 8}{a + 8} \cdot \frac{a \cdot a^{2} + a (- 9 a) + 8 a}{2 a^{2} - 128}$

Step 5.3

Factor $a$ out of $8 a$ .

$\frac{a - 8}{a + 8} \cdot \frac{a \cdot a^{2} + a (- 9 a) + a \cdot 8}{2 a^{2} - 128}$

Step 5.4

Factor $a$ out of $a \cdot a^{2} + a (- 9 a)$ .

$\frac{a - 8}{a + 8} \cdot \frac{a (a^{2} - 9 a) + a \cdot 8}{2 a^{2} - 128}$

Step 5.5

Factor $a$ out of $a (a^{2} - 9 a) + a \cdot 8$ .

$\frac{a - 8}{a + 8} \cdot \frac{a (a^{2} - 9 a + 8)}{2 a^{2} - 128}$

Step 6

Factor.

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Step 6.1

Factor $a^{2} - 9 a + 8$ using the AC method.

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Step 6.1.1

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 $8$ and whose sum is $- 9$ .

$- 8, - 1$

Step 6.1.2

Write the factored form using these integers.

$\frac{a - 8}{a + 8} \cdot \frac{a ((a - 8) (a - 1))}{2 a^{2} - 128}$

Step 6.2

Remove unnecessary parentheses.

$\frac{a - 8}{a + 8} \cdot \frac{a (a - 8) (a - 1)}{2 a^{2} - 128}$

Step 7

Factor $2$ out of $2 a^{2} - 128$ .

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Step 7.1

Factor $2$ out of $2 a^{2}$ .

$\frac{a - 8}{a + 8} \cdot \frac{a (a - 8) (a - 1)}{2 (a^{2}) - 128}$

Step 7.2

Factor $2$ out of $- 128$ .

$\frac{a - 8}{a + 8} \cdot \frac{a (a - 8) (a - 1)}{2 a^{2} + 2 \cdot - 64}$

Step 7.3

Factor $2$ out of $2 a^{2} + 2 \cdot - 64$ .

$\frac{a - 8}{a + 8} \cdot \frac{a (a - 8) (a - 1)}{2 (a^{2} - 64)}$

Step 8

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

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

Step 9

Factor.

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Step 9.1

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

$\frac{a - 8}{a + 8} \cdot \frac{a (a - 8) (a - 1)}{2 ((a + 8) (a - 8))}$

Step 9.2

Remove unnecessary parentheses.

$\frac{a - 8}{a + 8} \cdot \frac{a (a - 8) (a - 1)}{2 (a + 8) (a - 8)}$

Step 10

Reduce the expression $\frac{a (a - 8) (a - 1)}{2 (a + 8) (a - 8)}$ by cancelling the common factors.

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Step 10.1

Cancel the common factor.

$\frac{a - 8}{a + 8} \cdot \frac{a (a - 8) (a - 1)}{2 (a + 8) (a - 8)}$

Step 10.2

Rewrite the expression.

$\frac{a - 8}{a + 8} \cdot \frac{a (a - 1)}{2 (a + 8)}$