Solve 2m/3m-2-3/9m^2-4m/3m+2

$$\frac{ 2m }{ 3m-2 } - \frac{ 3 }{ 9 { m }^{ 2 } -4 } \frac{ m }{ 3m+2 }$$

Evaluate

$$\frac{m\left(18m^{2}+24m+5\right)}{\left(3m-2\right)\left(3m+2\right)^{2}}$$

Short Solution Steps

Multiply \frac{3}{9m^{2}-4} times \frac{m}{3m+2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{2m}{3m-2}-\frac{3m}{\left(9m^{2}-4\right)\left(3m+2\right)}$$

Use the distributive property to multiply 9m^{2}-4 by 3m+2.

$$\frac{2m}{3m-2}-\frac{3m}{27m^{3}+18m^{2}-12m-8}$$

Factor 27m^{3}+18m^{2}-12m-8.

$$\frac{2m}{3m-2}-\frac{3m}{\left(3m-2\right)\left(3m+2\right)^{2}}$$

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

$$\frac{2m\left(3m+2\right)^{2}}{\left(3m-2\right)\left(3m+2\right)^{2}}-\frac{3m}{\left(3m-2\right)\left(3m+2\right)^{2}}$$

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

$$\frac{2m\left(3m+2\right)^{2}-3m}{\left(3m-2\right)\left(3m+2\right)^{2}}$$

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

$$\frac{18m^{3}+24m^{2}+8m-3m}{\left(3m-2\right)\left(3m+2\right)^{2}}$$

Combine like terms in 18m^{3}+24m^{2}+8m-3m.

$$\frac{18m^{3}+24m^{2}+5m}{\left(3m-2\right)\left(3m+2\right)^{2}}$$

Expand \left(3m-2\right)\left(3m+2\right)^{2}.

$$\frac{18m^{3}+24m^{2}+5m}{27m^{3}+18m^{2}-12m-8}$$

Factor

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