Factor b^4-81

$b^{4} - 81$

Step 1

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

${(b^{2})}^{2} - 81$

Step 2

Rewrite $81$ as $9^{2}$ .

${(b^{2})}^{2} - 9^{2}$

Step 3

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

$(b^{2} + 9) (b^{2} - 9)$

Step 4

Simplify.

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

$(b^{2} + 9) (b^{2} - 3^{2})$

Factor.

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

$(b^{2} + 9) ((b + 3) (b - 3))$

Remove unnecessary parentheses.

$(b^{2} + 9) (b + 3) (b - 3)$

Factor b^4-81

Related Questions

Support LCMGCF.com by disabling your adblocker.