Algebra Examples

$6 x^{3} - 9 x - x^{5}$

Step 1

Factor $- x$ out of $6 x^{3} - 9 x - x^{5}$ .

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Step 1.1

Reorder the expression.

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Step 1.1.1

Move $- 9 x$ .

$6 x^{3} - x^{5} - 9 x$

Step 1.1.2

Reorder $6 x^{3}$ and $- x^{5}$ .

$- x^{5} + 6 x^{3} - 9 x$

Step 1.2

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

$- x (x^{4}) + 6 x^{3} - 9 x$

Step 1.3

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

$- x (x^{4}) - x (- 6 x^{2}) - 9 x$

Step 1.4

Factor $- x$ out of $- 9 x$ .

$- x (x^{4}) - x (- 6 x^{2}) - x (9)$

Step 1.5

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

$- x (x^{4} - 6 x^{2}) - x (9)$

Step 1.6

Factor $- x$ out of $- x (x^{4} - 6 x^{2}) - x (9)$ .

$- x (x^{4} - 6 x^{2} + 9)$

Step 2

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

$- x ({(x^{2})}^{2} - 6 x^{2} + 9)$

Step 3

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

$- x (u^{2} - 6 u + 9)$

Step 4

Factor using the perfect square rule.

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Step 4.1

Rewrite $9$ as $3^{2}$ .

$- x (u^{2} - 6 u + 3^{2})$

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.

$6 u = 2 \cdot u \cdot 3$

Step 4.3

Rewrite the polynomial.

$- x (u^{2} - 2 \cdot u \cdot 3 + 3^{2})$

Step 4.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 3$ .

$- x {(u - 3)}^{2}$

Step 5

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

$- x {(x^{2} - 3)}^{2}$