Solve n^2+2nc/n+3*5n+15/n^2-4c^2*n-2c/n=

$$\frac { n ^ { 2 } + 2 n c } { n + 3 } \cdot \frac { 5 n + 15 } { n ^ { 2 } - 4 c ^ { 2 } } \cdot \frac { n - 2 c } { n } =$$

Evaluate

$$5$$

Short Solution Steps

Multiply \frac{n^{2}+2nc}{n+3} times \frac{5n+15}{n^{2}-4c^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(n^{2}+2nc\right)\left(5n+15\right)}{\left(n+3\right)\left(n^{2}-4c^{2}\right)}\times \frac{n-2c}{n}$$

Multiply \frac{\left(n^{2}+2nc\right)\left(5n+15\right)}{\left(n+3\right)\left(n^{2}-4c^{2}\right)} times \frac{n-2c}{n} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(n^{2}+2nc\right)\left(5n+15\right)\left(n-2c\right)}{\left(n+3\right)\left(n^{2}-4c^{2}\right)n}$$

Factor the expressions that are not already factored.

$$\frac{5n\left(n+3\right)\left(n-2c\right)\left(n+2c\right)}{n\left(n+3\right)\left(n-2c\right)\left(n+2c\right)}$$

Cancel out n\left(n+3\right)\left(n-2c\right)\left(n+2c\right) in both numerator and denominator.

$$5$$

Factor

$$5$$