Factor 81m^8+2m^4+1

$81 m^{8} + 2 m^{4} + 1$

Rewrite the middle term.

$81 m^{8} + 2 (9 m^{4}) \cdot 1 - 16 m^{4} + 1$

Rearrange terms.

$81 m^{8} + 2 (9 m^{4}) \cdot 1 + 1 - 16 m^{4}$

Factor first three terms by perfect square rule.

${(9 m^{4} + 1)}^{2} - 16 m^{4}$

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

${(9 m^{4} + 1)}^{2} - {(4 m^{2})}^{2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 9 m^{4} + 1$ and $b = 4 m^{2}$ .

$(9 m^{4} + 1 + 4 m^{2}) (9 m^{4} + 1 - (4 m^{2}))$

Simplify.

Reorder terms.

$(9 m^{4} + 4 m^{2} + 1) (9 m^{4} + 1 - (4 m^{2}))$

Multiply $4$ by $- 1$ .

$(9 m^{4} + 4 m^{2} + 1) (9 m^{4} + 1 - 4 m^{2})$

Reorder terms.

$(9 m^{4} + 4 m^{2} + 1) (9 m^{4} - 4 m^{2} + 1)$

Factor 81m^8+2m^4+1

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