Factor 1/25+(25x^4)/36-(x^2)/3

$\frac{1}{25} + \frac{25 x^{4}}{36} - \frac{x^{2}}{3}$

Rearrange terms.

$\frac{1}{25} - \frac{x^{2}}{3} + \frac{25 x^{4}}{36}$

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

$\frac{1^{2}}{25} - \frac{x^{2}}{3} + \frac{25 x^{4}}{36}$

Rewrite $25$ as $5^{2}$ .

$\frac{1^{2}}{5^{2}} - \frac{x^{2}}{3} + \frac{25 x^{4}}{36}$

Rewrite $\frac{1^{2}}{5^{2}}$ as ${(\frac{1}{5})}^{2}$ .

${(\frac{1}{5})}^{2} - \frac{x^{2}}{3} + \frac{25 x^{4}}{36}$

Rewrite $25 x^{4}$ as ${(5 x^{2})}^{2}$ .

${(\frac{1}{5})}^{2} - \frac{x^{2}}{3} + \frac{{(5 x^{2})}^{2}}{36}$

Rewrite $36$ as $6^{2}$ .

${(\frac{1}{5})}^{2} - \frac{x^{2}}{3} + \frac{{(5 x^{2})}^{2}}{6^{2}}$

Rewrite $\frac{{(5 x^{2})}^{2}}{6^{2}}$ as ${(\frac{5 x^{2}}{6})}^{2}$ .

${(\frac{1}{5})}^{2} - \frac{x^{2}}{3} + {(\frac{5 x^{2}}{6})}^{2}$

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

$\frac{x^{2}}{3} = 2 \cdot \frac{1}{5} \cdot \frac{5 x^{2}}{6}$

Rewrite the polynomial.

${(\frac{1}{5})}^{2} - 2 \cdot \frac{1}{5} \cdot \frac{5 x^{2}}{6} + {(\frac{5 x^{2}}{6})}^{2}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = \frac{1}{5}$ and $b = \frac{5 x^{2}}{6}$ .

${(\frac{1}{5} - \frac{5 x^{2}}{6})}^{2}$