Factor m^4n^3+24m^2n^3+144n^3

$m^{4} n^{3} + 24 m^{2} n^{3} + 144 n^{3}$

Step 1

Factor $n^{3}$ out of $m^{4} n^{3} + 24 m^{2} n^{3} + 144 n^{3}$ .

Factor $n^{3}$ out of $m^{4} n^{3}$ .

$n^{3} m^{4} + 24 m^{2} n^{3} + 144 n^{3}$

Factor $n^{3}$ out of $24 m^{2} n^{3}$ .

$n^{3} m^{4} + n^{3} (24 m^{2}) + 144 n^{3}$

Factor $n^{3}$ out of $144 n^{3}$ .

$n^{3} m^{4} + n^{3} (24 m^{2}) + n^{3} \cdot 144$

Factor $n^{3}$ out of $n^{3} m^{4} + n^{3} (24 m^{2})$ .

$n^{3} (m^{4} + 24 m^{2}) + n^{3} \cdot 144$

Factor $n^{3}$ out of $n^{3} (m^{4} + 24 m^{2}) + n^{3} \cdot 144$ .

$n^{3} (m^{4} + 24 m^{2} + 144)$

Step 2

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

$n^{3} ({(m^{2})}^{2} + 24 m^{2} + 144)$

Step 3

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

$n^{3} (u^{2} + 24 u + 144)$

Step 4

Factor using the perfect square rule.

Rewrite $144$ as $12^{2}$ .

$n^{3} (u^{2} + 24 u + 12^{2})$

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

$24 u = 2 \cdot u \cdot 12$

Rewrite the polynomial.

$n^{3} (u^{2} + 2 \cdot u \cdot 12 + 12^{2})$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 12$ .

$n^{3} {(u + 12)}^{2}$

Step 5

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

$n^{3} {(m^{2} + 12)}^{2}$