Solve w^2-2w/3+1/9

$$w ^ { 2 } - \frac { 2 w } { 3 } + \frac { 1 } { 9 }$$

Evaluate

$$\frac{\left(3w-1\right)^{2}}{9}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 9 is 9. Multiply \frac{2w}{3} times \frac{3}{3}.

$$w^{2}-\frac{3\times 2w}{9}+\frac{1}{9}$$

Since -\frac{3\times 2w}{9} and \frac{1}{9} have the same denominator, add them by adding their numerators.

$$w^{2}+\frac{-3\times 2w+1}{9}$$

Do the multiplications in -3\times 2w+1.

$$w^{2}+\frac{-6w+1}{9}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply w^{2} times \frac{9}{9}.

$$\frac{9w^{2}}{9}+\frac{-6w+1}{9}$$

Since \frac{9w^{2}}{9} and \frac{-6w+1}{9} have the same denominator, add them by adding their numerators.

$$\frac{9w^{2}-6w+1}{9}$$

Factor

$$\frac{\left(3w-1\right)^{2}}{9}$$

Solution Steps

Factor out \frac{1}{9}.

$$\frac{9w^{2}-6w+1}{9}$$

Consider 9w^{2}-6w+1. Use the perfect square formula, a^{2}-2ab+b^{2}=\left(a-b\right)^{2}, where a=3w and b=1.

$$\left(3w-1\right)^{2}$$

Rewrite the complete factored expression.

$$\frac{\left(3w-1\right)^{2}}{9}$$