Factor 16y^4-81

$16 y^{4} - 81$

Step 1

Rewrite $16 y^{4}$ as ${(4 y^{2})}^{2}$ .

${(4 y^{2})}^{2} - 81$

Step 2

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

${(4 y^{2})}^{2} - 9^{2}$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 y^{2}$ and $b = 9$ .

$(4 y^{2} + 9) (4 y^{2} - 9)$

Step 4

Simplify.

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

$(4 y^{2} + 9) ({(2 y)}^{2} - 9)$

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

$(4 y^{2} + 9) ({(2 y)}^{2} - 3^{2})$

Factor.

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

$(4 y^{2} + 9) ((2 y + 3) (2 y - 3))$

Remove unnecessary parentheses.

$(4 y^{2} + 9) (2 y + 3) (2 y - 3)$

Factor 16y^4-81

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