Solve 1+2b/3+b^2/9=

$$1 + \frac { 2 b } { 3 } + \frac { b ^ { 2 } } { 9 } =$$

Evaluate

$$\frac{\left(b+3\right)^{2}}{9}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

$$\frac{3}{3}+\frac{2b}{3}+\frac{b^{2}}{9}$$

Since \frac{3}{3} and \frac{2b}{3} have the same denominator, add them by adding their numerators.

$$\frac{3+2b}{3}+\frac{b^{2}}{9}$$

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

$$\frac{3\left(3+2b\right)}{9}+\frac{b^{2}}{9}$$

Since \frac{3\left(3+2b\right)}{9} and \frac{b^{2}}{9} have the same denominator, add them by adding their numerators.

$$\frac{3\left(3+2b\right)+b^{2}}{9}$$

Do the multiplications in 3\left(3+2b\right)+b^{2}.

$$\frac{9+6b+b^{2}}{9}$$

Factor

$$\frac{\left(b+3\right)^{2}}{9}$$

Solution Steps

Factor out \frac{1}{9}.

$$\frac{9+6b+b^{2}}{9}$$

Consider 9+6b+b^{2}. Use the perfect square formula, p^{2}+2pq+q^{2}=\left(p+q\right)^{2}, where p=b and q=3.

$$\left(b+3\right)^{2}$$

Rewrite the complete factored expression.

$$\frac{\left(b+3\right)^{2}}{9}$$