Factor 16m^6-2m^3n^2+(n^4)/16

$16 m^{6} - 2 m^{3} n^{2} + \frac{n^{4}}{16}$

Step 1

Rewrite $16 m^{6}$ as ${(4 m^{3})}^{2}$ .

${(4 m^{3})}^{2} - 2 m^{3} n^{2} + \frac{n^{4}}{16}$

Step 2

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

${(4 m^{3})}^{2} - 2 m^{3} n^{2} + \frac{{(n^{2})}^{2}}{16}$

Step 3

Rewrite $16$ as $4^{2}$ .

${(4 m^{3})}^{2} - 2 m^{3} n^{2} + \frac{{(n^{2})}^{2}}{4^{2}}$

Step 4

Rewrite $\frac{{(n^{2})}^{2}}{4^{2}}$ as ${(\frac{n^{2}}{4})}^{2}$ .

${(4 m^{3})}^{2} - 2 m^{3} n^{2} + {(\frac{n^{2}}{4})}^{2}$

Step 5

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

$2 m^{3} n^{2} = 2 \cdot (4 m^{3}) \cdot \frac{n^{2}}{4}$

Step 6

Rewrite the polynomial.

${(4 m^{3})}^{2} - 2 \cdot (4 m^{3}) \cdot \frac{n^{2}}{4} + {(\frac{n^{2}}{4})}^{2}$

Step 7

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 4 m^{3}$ and $b = \frac{n^{2}}{4}$ .

${(4 m^{3} - \frac{n^{2}}{4})}^{2}$