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

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

Step 1

Factor using the perfect square rule.

Rearrange terms.

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

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

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

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

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

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

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

Rewrite $\frac{25}{36} b^{4}$ as ${(\frac{5}{6} b^{2})}^{2}$ .

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

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

$\frac{b^{2}}{3} = 2 \cdot \frac{1}{5} \cdot (\frac{5}{6} b^{2})$

Rewrite the polynomial.

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

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

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

Step 2

Combine $b^{2}$ and $\frac{5}{6}$ .

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

Step 3

Move $5$ to the left of $b^{2}$ .

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

Step 4

Multiply $5$ by $b^{2}$ .

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