Factor 36m^6+12m^4+m^2

$36 m^{6} + 12 m^{4} + m^{2}$

Factor $m^{2}$ out of $36 m^{6} + 12 m^{4} + m^{2}$ .

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Factor $m^{2}$ out of $36 m^{6}$ .

$m^{2} (36 m^{4}) + 12 m^{4} + m^{2}$

Factor $m^{2}$ out of $12 m^{4}$ .

$m^{2} (36 m^{4}) + m^{2} (12 m^{2}) + m^{2}$

Multiply by $1$ .

$m^{2} (36 m^{4}) + m^{2} (12 m^{2}) + m^{2} \cdot 1$

Factor $m^{2}$ out of $m^{2} (36 m^{4}) + m^{2} (12 m^{2})$ .

$m^{2} (36 m^{4} + 12 m^{2}) + m^{2} \cdot 1$

Factor $m^{2}$ out of $m^{2} (36 m^{4} + 12 m^{2}) + m^{2} \cdot 1$ .

$m^{2} (36 m^{4} + 12 m^{2} + 1)$

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

$m^{2} (36 {(m^{2})}^{2} + 12 m^{2} + 1)$

Let $u = m^{2}$ . Substitute $u$ for all occurrences of $m^{2}$ .

$m^{2} (36 u^{2} + 12 u + 1)$

Factor using the perfect square rule.

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Rewrite $36 u^{2}$ as ${(6 u)}^{2}$ .

$m^{2} ({(6 u)}^{2} + 12 u + 1)$

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

$m^{2} ({(6 u)}^{2} + 12 u + 1^{2})$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 u = 2 \cdot (6 u) \cdot 1$

Rewrite the polynomial.

$m^{2} ({(6 u)}^{2} + 2 \cdot (6 u) \cdot 1 + 1^{2})$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 6 u$ and $b = 1$ .

$m^{2} {(6 u + 1)}^{2}$

Replace all occurrences of $u$ with $m^{2}$ .

$m^{2} {(6 m^{2} + 1)}^{2}$

Factor 36m^6+12m^4+m^2

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