Solve (frac{15k^{2}-7k-2}{k^{2}+8k}div3k-2/k^3+2k^2-48k)*k^2-3k+2/5k^2-9k-2=

$$( \frac { 15 k ^ { 2 } - 7 k - 2 } { k ^ { 2 } + 8 k } \div \frac { 3 k - 2 } { k ^ { 3 } + 2 k ^ { 2 } - 48 k } ) \cdot \frac { k ^ { 2 } - 3 k + 2 } { 5 k ^ { 2 } - 9 k - 2 } =$$

Evaluate

$$\left(k-6\right)\left(k-1\right)$$

Short Solution Steps

Divide \frac{15k^{2}-7k-2}{k^{2}+8k} by \frac{3k-2}{k^{3}+2k^{2}-48k} by multiplying \frac{15k^{2}-7k-2}{k^{2}+8k} by the reciprocal of \frac{3k-2}{k^{3}+2k^{2}-48k}.

$$\frac{\left(15k^{2}-7k-2\right)\left(k^{3}+2k^{2}-48k\right)}{\left(k^{2}+8k\right)\left(3k-2\right)}\times \frac{k^{2}-3k+2}{5k^{2}-9k-2}$$

Factor the expressions that are not already factored in \frac{\left(15k^{2}-7k-2\right)\left(k^{3}+2k^{2}-48k\right)}{\left(k^{2}+8k\right)\left(3k-2\right)}.

$$\frac{k\left(k-6\right)\left(3k-2\right)\left(k+8\right)\left(5k+1\right)}{k\left(3k-2\right)\left(k+8\right)}\times \frac{k^{2}-3k+2}{5k^{2}-9k-2}$$

Cancel out k\left(3k-2\right)\left(k+8\right) in both numerator and denominator.

$$\left(k-6\right)\left(5k+1\right)\times \frac{k^{2}-3k+2}{5k^{2}-9k-2}$$

Expand the expression.

$$\left(5k^{2}-29k-6\right)\times \frac{k^{2}-3k+2}{5k^{2}-9k-2}$$

Factor the expressions that are not already factored in \frac{k^{2}-3k+2}{5k^{2}-9k-2}.

$$\left(5k^{2}-29k-6\right)\times \frac{\left(k-2\right)\left(k-1\right)}{\left(k-2\right)\left(5k+1\right)}$$

Cancel out k-2 in both numerator and denominator.

$$\left(5k^{2}-29k-6\right)\times \frac{k-1}{5k+1}$$

Express \left(5k^{2}-29k-6\right)\times \frac{k-1}{5k+1} as a single fraction.

$$\frac{\left(5k^{2}-29k-6\right)\left(k-1\right)}{5k+1}$$

Factor the expressions that are not already factored.

$$\frac{\left(k-6\right)\left(k-1\right)\left(5k+1\right)}{5k+1}$$

Cancel out 5k+1 in both numerator and denominator.

$$\left(k-6\right)\left(k-1\right)$$

Expand the expression.

$$k^{2}-7k+6$$

Expand

$$k^{2}-7k+6$$

Short Solution Steps

Divide \frac{15k^{2}-7k-2}{k^{2}+8k} by \frac{3k-2}{k^{3}+2k^{2}-48k} by multiplying \frac{15k^{2}-7k-2}{k^{2}+8k} by the reciprocal of \frac{3k-2}{k^{3}+2k^{2}-48k}.

$$\frac{\left(15k^{2}-7k-2\right)\left(k^{3}+2k^{2}-48k\right)}{\left(k^{2}+8k\right)\left(3k-2\right)}\times \frac{k^{2}-3k+2}{5k^{2}-9k-2}$$

Factor the expressions that are not already factored in \frac{\left(15k^{2}-7k-2\right)\left(k^{3}+2k^{2}-48k\right)}{\left(k^{2}+8k\right)\left(3k-2\right)}.

$$\frac{k\left(k-6\right)\left(3k-2\right)\left(k+8\right)\left(5k+1\right)}{k\left(3k-2\right)\left(k+8\right)}\times \frac{k^{2}-3k+2}{5k^{2}-9k-2}$$

Cancel out k\left(3k-2\right)\left(k+8\right) in both numerator and denominator.

$$\left(k-6\right)\left(5k+1\right)\times \frac{k^{2}-3k+2}{5k^{2}-9k-2}$$

Expand the expression.

$$\left(5k^{2}-29k-6\right)\times \frac{k^{2}-3k+2}{5k^{2}-9k-2}$$

Factor the expressions that are not already factored in \frac{k^{2}-3k+2}{5k^{2}-9k-2}.

$$\left(5k^{2}-29k-6\right)\times \frac{\left(k-2\right)\left(k-1\right)}{\left(k-2\right)\left(5k+1\right)}$$

Cancel out k-2 in both numerator and denominator.

$$\left(5k^{2}-29k-6\right)\times \frac{k-1}{5k+1}$$

Express \left(5k^{2}-29k-6\right)\times \frac{k-1}{5k+1} as a single fraction.

$$\frac{\left(5k^{2}-29k-6\right)\left(k-1\right)}{5k+1}$$

Factor the expressions that are not already factored.

$$\frac{\left(k-6\right)\left(k-1\right)\left(5k+1\right)}{5k+1}$$

Cancel out 5k+1 in both numerator and denominator.

$$\left(k-6\right)\left(k-1\right)$$

Expand the expression.

$$k^{2}-7k+6$$