Finite Math Examples

$a^{5} - 8 a^{3} + 16 a$

Step 1

Factor $a$ out of $a^{5} - 8 a^{3} + 16 a$ .

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Step 1.1

Factor $a$ out of $a^{5}$ .

$a \cdot a^{4} - 8 a^{3} + 16 a$

Step 1.2

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

$a \cdot a^{4} + a (- 8 a^{2}) + 16 a$

Step 1.3

Factor $a$ out of $16 a$ .

$a \cdot a^{4} + a (- 8 a^{2}) + a \cdot 16$

Step 1.4

Factor $a$ out of $a \cdot a^{4} + a (- 8 a^{2})$ .

$a (a^{4} - 8 a^{2}) + a \cdot 16$

Step 1.5

Factor $a$ out of $a (a^{4} - 8 a^{2}) + a \cdot 16$ .

$a (a^{4} - 8 a^{2} + 16)$

Step 2

Rewrite $a^{4}$ as ${(a^{2})}^{2}$ .

$a ({(a^{2})}^{2} - 8 a^{2} + 16)$

Step 3

Let $u = a^{2}$ . Substitute $u$ for all occurrences of $a^{2}$ .

$a (u^{2} - 8 u + 16)$

Step 4

Factor using the perfect square rule.

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Step 4.1

Rewrite $16$ as $4^{2}$ .

$a (u^{2} - 8 u + 4^{2})$

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.

$8 u = 2 \cdot u \cdot 4$

Step 4.3

Rewrite the polynomial.

$a (u^{2} - 2 \cdot u \cdot 4 + 4^{2})$

Step 4.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 4$ .

$a {(u - 4)}^{2}$

Step 5

Replace all occurrences of $u$ with $a^{2}$ .

$a {(a^{2} - 4)}^{2}$

Step 6

Rewrite $4$ as $2^{2}$ .

$a {(a^{2} - 2^{2})}^{2}$

Step 7

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 $b = 2$ .

$a {((a + 2) (a - 2))}^{2}$

Step 8

Factor.

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Step 8.1

Apply the product rule to $(a + 2) (a - 2)$ .

$a ({(a + 2)}^{2} {(a - 2)}^{2})$

Step 8.2

Remove unnecessary parentheses.

$a {(a + 2)}^{2} {(a - 2)}^{2}$