Factor 81x^4-1

$81 x^{4} - 1$

Step 1

Rewrite $81 x^{4}$ as ${(9 x^{2})}^{2}$ .

${(9 x^{2})}^{2} - 1$

Step 2

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

${(9 x^{2})}^{2} - 1^{2}$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 9 x^{2}$ and $b = 1$ .

$(9 x^{2} + 1) (9 x^{2} - 1)$

Step 4

Simplify.

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$(9 x^{2} + 1) ({(3 x)}^{2} - 1)$

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

$(9 x^{2} + 1) ({(3 x)}^{2} - 1^{2})$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 1$ .

$(9 x^{2} + 1) ((3 x + 1) (3 x - 1))$

Remove unnecessary parentheses.

$(9 x^{2} + 1) (3 x + 1) (3 x - 1)$

Factor 81x^4-1

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