Solve 1/2(3+h)^2-1/2(3)^2/h

$$\frac { \frac { 1 } { 2 ( 3 + h ) ^ { 2 } } - \frac { 1 } { 2 ( 3 ) ^ { 2 } } } { h }$$

Evaluate

$$-\frac{h+6}{18\left(h+3\right)^{2}}$$

Short Solution Steps

Calculate 3 to the power of 2 and get 9.

$$\frac{\frac{1}{2\left(3+h\right)^{2}}-\frac{1}{2\times 9}}{h}$$

Multiply 2 and 9 to get 18.

$$\frac{\frac{1}{2\left(3+h\right)^{2}}-\frac{1}{18}}{h}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(3+h\right)^{2} and 18 is 18\left(h+3\right)^{2}. Multiply \frac{1}{2\left(3+h\right)^{2}} times \frac{9}{9}. Multiply \frac{1}{18} times \frac{\left(h+3\right)^{2}}{\left(h+3\right)^{2}}.

$$\frac{\frac{9}{18\left(h+3\right)^{2}}-\frac{\left(h+3\right)^{2}}{18\left(h+3\right)^{2}}}{h}$$

Since \frac{9}{18\left(h+3\right)^{2}} and \frac{\left(h+3\right)^{2}}{18\left(h+3\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{9-\left(h+3\right)^{2}}{18\left(h+3\right)^{2}}}{h}$$

Do the multiplications in 9-\left(h+3\right)^{2}.

$$\frac{\frac{9-h^{2}-6h-9}{18\left(h+3\right)^{2}}}{h}$$

Combine like terms in 9-h^{2}-6h-9.

$$\frac{\frac{-h^{2}-6h}{18\left(h+3\right)^{2}}}{h}$$

Express \frac{\frac{-h^{2}-6h}{18\left(h+3\right)^{2}}}{h} as a single fraction.

$$\frac{-h^{2}-6h}{18\left(h+3\right)^{2}h}$$

Factor the expressions that are not already factored.

$$\frac{h\left(-h-6\right)}{18h\left(h+3\right)^{2}}$$

Cancel out h in both numerator and denominator.

$$\frac{-h-6}{18\left(h+3\right)^{2}}$$

Expand the expression.

$$\frac{-h-6}{18h^{2}+108h+162}$$

Expand

$$-\frac{h+6}{18\left(h+3\right)^{2}}$$

Short Solution Steps

Calculate 3 to the power of 2 and get 9.

$$\frac{\frac{1}{2\left(3+h\right)^{2}}-\frac{1}{2\times 9}}{h}$$

Multiply 2 and 9 to get 18.

$$\frac{\frac{1}{2\left(3+h\right)^{2}}-\frac{1}{18}}{h}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(3+h\right)^{2} and 18 is 18\left(h+3\right)^{2}. Multiply \frac{1}{2\left(3+h\right)^{2}} times \frac{9}{9}. Multiply \frac{1}{18} times \frac{\left(h+3\right)^{2}}{\left(h+3\right)^{2}}.

$$\frac{\frac{9}{18\left(h+3\right)^{2}}-\frac{\left(h+3\right)^{2}}{18\left(h+3\right)^{2}}}{h}$$

Since \frac{9}{18\left(h+3\right)^{2}} and \frac{\left(h+3\right)^{2}}{18\left(h+3\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{9-\left(h+3\right)^{2}}{18\left(h+3\right)^{2}}}{h}$$

Do the multiplications in 9-\left(h+3\right)^{2}.

$$\frac{\frac{9-h^{2}-6h-9}{18\left(h+3\right)^{2}}}{h}$$

Combine like terms in 9-h^{2}-6h-9.

$$\frac{\frac{-h^{2}-6h}{18\left(h+3\right)^{2}}}{h}$$

Express \frac{\frac{-h^{2}-6h}{18\left(h+3\right)^{2}}}{h} as a single fraction.

$$\frac{-h^{2}-6h}{18\left(h+3\right)^{2}h}$$

Factor the expressions that are not already factored.

$$\frac{h\left(-h-6\right)}{18h\left(h+3\right)^{2}}$$

Cancel out h in both numerator and denominator.

$$\frac{-h-6}{18\left(h+3\right)^{2}}$$

Expand the expression.

$$\frac{-h-6}{18h^{2}+108h+162}$$