Factor 16x^8y^4-81z^4

$16 x^{8} y^{4} - 81 z^{4}$

Step 1

Rewrite $16 x^{8} y^{4}$ as ${(4 x^{4} y^{2})}^{2}$ .

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

Step 2

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

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

Step 3

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

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

Step 4

Simplify.

Remove parentheses.

$(4 x^{4} y^{2} + 9 z^{2}) (4 x^{4} y^{2} - 1 \cdot 9 z^{2})$

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

$(4 x^{4} y^{2} + 9 z^{2}) ({(2 x^{2} y)}^{2} - 1 \cdot 9 z^{2})$

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

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

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

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

Factor.

Multiply $3$ by $- 1$ .

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

Remove unnecessary parentheses.

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

Factor 16x^8y^4-81z^4

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