Factor -5x^3+6x-x^5

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

Step 1

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

Reorder the expression.

Move $6 x$ .

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

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

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

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

$- x (x^{4}) - 5 x^{3} + 6 x$

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

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

Factor $- x$ out of $6 x$ .

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

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

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

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

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

Step 2

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

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

Step 3

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

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

Step 4

Factor $u^{2} + 5 u - 6$ using the AC method.

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 $- 6$ and whose sum is $5$ .

$- 1, 6$

Write the factored form using these integers.

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

Step 5

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

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

Step 6

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

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

Step 7

Factor.

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 $b = 1$ .

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

Remove unnecessary parentheses.

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

Factor -5x^3+6x-x^5

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