Finite Math Examples

$6 x^{5} - 30 x^{3} + 24 x$

Step 1

Factor $6 x$ out of $6 x^{5} - 30 x^{3} + 24 x$ .

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Step 1.1

Factor $6 x$ out of $6 x^{5}$ .

$6 x (x^{4}) - 30 x^{3} + 24 x$

Step 1.2

Factor $6 x$ out of $- 30 x^{3}$ .

$6 x (x^{4}) + 6 x (- 5 x^{2}) + 24 x$

Step 1.3

Factor $6 x$ out of $24 x$ .

$6 x (x^{4}) + 6 x (- 5 x^{2}) + 6 x (4)$

Step 1.4

Factor $6 x$ out of $6 x (x^{4}) + 6 x (- 5 x^{2})$ .

$6 x (x^{4} - 5 x^{2}) + 6 x (4)$

Step 1.5

Factor $6 x$ out of $6 x (x^{4} - 5 x^{2}) + 6 x (4)$ .

$6 x (x^{4} - 5 x^{2} + 4)$

Step 2

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

$6 x ({(x^{2})}^{2} - 5 x^{2} + 4)$

Step 3

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

$6 x (u^{2} - 5 u + 4)$

Step 4

Factor $u^{2} - 5 u + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Step 4.2

Write the factored form using these integers.

$6 x ((u - 4) (u - 1))$

Step 5

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

$6 x ((x^{2} - 4) (x^{2} - 1))$

Step 6

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

$6 x ((x^{2} - 2^{2}) (x^{2} - 1))$

Step 7

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

$6 x ((x + 2) (x - 2) (x^{2} - 1))$

Step 8

Rewrite $1$ as $1^{2}$ .

$6 x ((x + 2) (x - 2) (x^{2} - 1^{2}))$

Step 9

Factor.

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Step 9.1

Factor.

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Step 9.1.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$6 x ((x + 2) (x - 2) ((x + 1) (x - 1)))$

Step 9.1.2

Remove unnecessary parentheses.

$6 x ((x + 2) (x - 2) (x + 1) (x - 1))$

Step 9.2

Remove unnecessary parentheses.

$6 x (x + 2) (x - 2) (x + 1) (x - 1)$