Solve (2x+1)^2/5-(x+3)(x-3)/3*20/3

$$\frac { ( 2 x + 1 ) ^ { 2 } } { 5 } - \frac { ( x + 3 ) ( x - 3 ) } { 3 } \cdot \frac { 20 } { 3 }$$

Evaluate

$$-\frac{64x^{2}}{45}+\frac{4x}{5}+\frac{101}{5}$$

Short Solution Steps

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\frac{\left(2x+1\right)^{2}}{5}-\frac{x^{2}-9}{3}\times \frac{20}{3}$$

Multiply \frac{x^{2}-9}{3} times \frac{20}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(2x+1\right)^{2}}{5}-\frac{\left(x^{2}-9\right)\times 20}{3\times 3}$$

Multiply 3 and 3 to get 9.

$$\frac{\left(2x+1\right)^{2}}{5}-\frac{\left(x^{2}-9\right)\times 20}{9}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5 and 9 is 45. Multiply \frac{\left(2x+1\right)^{2}}{5} times \frac{9}{9}. Multiply \frac{\left(x^{2}-9\right)\times 20}{9} times \frac{5}{5}.

$$\frac{9\left(2x+1\right)^{2}}{45}-\frac{5\left(x^{2}-9\right)\times 20}{45}$$

Since \frac{9\left(2x+1\right)^{2}}{45} and \frac{5\left(x^{2}-9\right)\times 20}{45} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9\left(2x+1\right)^{2}-5\left(x^{2}-9\right)\times 20}{45}$$

Do the multiplications in 9\left(2x+1\right)^{2}-5\left(x^{2}-9\right)\times 20.

$$\frac{36x^{2}+36x+9-100x^{2}+900}{45}$$

Combine like terms in 36x^{2}+36x+9-100x^{2}+900.

$$\frac{-64x^{2}+36x+909}{45}$$

Expand

$$-\frac{64x^{2}}{45}+\frac{4x}{5}+\frac{101}{5}$$

Short Solution Steps

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\frac{\left(2x+1\right)^{2}}{5}-\frac{x^{2}-9}{3}\times \frac{20}{3}$$

Multiply \frac{x^{2}-9}{3} times \frac{20}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(2x+1\right)^{2}}{5}-\frac{\left(x^{2}-9\right)\times 20}{3\times 3}$$

Multiply 3 and 3 to get 9.

$$\frac{\left(2x+1\right)^{2}}{5}-\frac{\left(x^{2}-9\right)\times 20}{9}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5 and 9 is 45. Multiply \frac{\left(2x+1\right)^{2}}{5} times \frac{9}{9}. Multiply \frac{\left(x^{2}-9\right)\times 20}{9} times \frac{5}{5}.

$$\frac{9\left(2x+1\right)^{2}}{45}-\frac{5\left(x^{2}-9\right)\times 20}{45}$$

Since \frac{9\left(2x+1\right)^{2}}{45} and \frac{5\left(x^{2}-9\right)\times 20}{45} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9\left(2x+1\right)^{2}-5\left(x^{2}-9\right)\times 20}{45}$$

Do the multiplications in 9\left(2x+1\right)^{2}-5\left(x^{2}-9\right)\times 20.

$$\frac{36x^{2}+36x+9-100x^{2}+900}{45}$$

Combine like terms in 36x^{2}+36x+9-100x^{2}+900.

$$\frac{-64x^{2}+36x+909}{45}$$