Solve 10k^2-90k/20k^2+40kdivk^2-17k+72/16k^2+32k

$$\frac { 10 k ^ { 2 } - 90 k } { 20 k ^ { 2 } + 40 k } \div \frac { k ^ { 2 } - 17 k + 72 } { 16 k ^ { 2 } + 32 k }$$

Evaluate

$$\frac{8k}{k-8}$$

Short Solution Steps

Divide \frac{10k^{2}-90k}{20k^{2}+40k} by \frac{k^{2}-17k+72}{16k^{2}+32k} by multiplying \frac{10k^{2}-90k}{20k^{2}+40k} by the reciprocal of \frac{k^{2}-17k+72}{16k^{2}+32k}.

$$\frac{\left(10k^{2}-90k\right)\left(16k^{2}+32k\right)}{\left(20k^{2}+40k\right)\left(k^{2}-17k+72\right)}$$

Factor the expressions that are not already factored.

$$\frac{10\times 16\left(k-9\right)\left(k+2\right)k^{2}}{20k\left(k-9\right)\left(k-8\right)\left(k+2\right)}$$

Cancel out 2\times 10k\left(k-9\right)\left(k+2\right) in both numerator and denominator.

$$\frac{8k}{k-8}$$

Expand

$$\frac{8k}{k-8}$$

Short Solution Steps

Divide \frac{10k^{2}-90k}{20k^{2}+40k} by \frac{k^{2}-17k+72}{16k^{2}+32k} by multiplying \frac{10k^{2}-90k}{20k^{2}+40k} by the reciprocal of \frac{k^{2}-17k+72}{16k^{2}+32k}.

$$\frac{\left(10k^{2}-90k\right)\left(16k^{2}+32k\right)}{\left(20k^{2}+40k\right)\left(k^{2}-17k+72\right)}$$

Factor the expressions that are not already factored.

$$\frac{10\times 16\left(k-9\right)\left(k+2\right)k^{2}}{20k\left(k-9\right)\left(k-8\right)\left(k+2\right)}$$

Cancel out 2\times 10k\left(k-9\right)\left(k+2\right) in both numerator and denominator.

$$\frac{8k}{k-8}$$