Solve frac{y^{2}-yc-2c^{2}}{10y+5c}*4y^2-c^2/3y-6cdivy^2-c^2/15y-15c

$$\frac { y ^ { 2 } - y c - 2 c ^ { 2 } } { 10 y + 5 c } \cdot \frac { 4 y ^ { 2 } - c ^ { 2 } } { 3 y - 6 c } \div \frac { y ^ { 2 } - c ^ { 2 } } { 15 y - 15 c }$$

Evaluate

$$2y-c$$

Short Solution Steps

Multiply \frac{y^{2}-yc-2c^{2}}{10y+5c} times \frac{4y^{2}-c^{2}}{3y-6c} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)}}{\frac{y^{2}-c^{2}}{15y-15c}}$$

Factor the expressions that are not already factored in \frac{y^{2}-c^{2}}{15y-15c}.

$$\frac{\frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)}}{\frac{\left(y+c\right)\left(y-c\right)}{15\left(y-c\right)}}$$

Cancel out y-c in both numerator and denominator.

$$\frac{\frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)}}{\frac{y+c}{15}}$$

Divide \frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)} by \frac{y+c}{15} by multiplying \frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)} by the reciprocal of \frac{y+c}{15}.

$$\frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)\times 15}{\left(10y+5c\right)\left(3y-6c\right)\left(y+c\right)}$$

Factor the expressions that are not already factored.

$$\frac{15\left(y+c\right)\left(2y+c\right)\left(y-2c\right)\left(2y-c\right)}{3\times 5\left(y+c\right)\left(2y+c\right)\left(y-2c\right)}$$

Cancel out 3\times 5\left(y+c\right)\left(2y+c\right)\left(y-2c\right) in both numerator and denominator.

$$2y-c$$

Expand

$$2y-c$$

Short Solution Steps

Multiply \frac{y^{2}-yc-2c^{2}}{10y+5c} times \frac{4y^{2}-c^{2}}{3y-6c} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)}}{\frac{y^{2}-c^{2}}{15y-15c}}$$

Factor the expressions that are not already factored in \frac{y^{2}-c^{2}}{15y-15c}.

$$\frac{\frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)}}{\frac{\left(y+c\right)\left(y-c\right)}{15\left(y-c\right)}}$$

Cancel out y-c in both numerator and denominator.

$$\frac{\frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)}}{\frac{y+c}{15}}$$

Divide \frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)} by \frac{y+c}{15} by multiplying \frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)}{\left(10y+5c\right)\left(3y-6c\right)} by the reciprocal of \frac{y+c}{15}.

$$\frac{\left(y^{2}-yc-2c^{2}\right)\left(4y^{2}-c^{2}\right)\times 15}{\left(10y+5c\right)\left(3y-6c\right)\left(y+c\right)}$$

Factor the expressions that are not already factored.

$$\frac{15\left(y+c\right)\left(2y+c\right)\left(y-2c\right)\left(2y-c\right)}{3\times 5\left(y+c\right)\left(2y+c\right)\left(y-2c\right)}$$

Cancel out 3\times 5\left(y+c\right)\left(2y+c\right)\left(y-2c\right) in both numerator and denominator.

$$2y-c$$