Solve frac{2x^{2}-9x-18}{4x^2-24x}div2x^3+7x^2/x^2-16*1/x

$$\frac { 2 x ^ { 2 } - 9 x - 18 } { 4 x ^ { 2 } - 24 x } \div \frac { 2 x ^ { 3 } + 7 x ^ { 2 } } { x ^ { 2 } - 16 } \times \frac { 1 } { x }$$

Evaluate

$$\frac{\left(2x+3\right)\left(x^{2}-16\right)}{4\left(2x+7\right)x^{4}}$$

Short Solution Steps

Divide \frac{2x^{2}-9x-18}{4x^{2}-24x} by \frac{2x^{3}+7x^{2}}{x^{2}-16} by multiplying \frac{2x^{2}-9x-18}{4x^{2}-24x} by the reciprocal of \frac{2x^{3}+7x^{2}}{x^{2}-16}.

$$\frac{\left(2x^{2}-9x-18\right)\left(x^{2}-16\right)}{\left(4x^{2}-24x\right)\left(2x^{3}+7x^{2}\right)}\times \frac{1}{x}$$

Factor the expressions that are not already factored in \frac{\left(2x^{2}-9x-18\right)\left(x^{2}-16\right)}{\left(4x^{2}-24x\right)\left(2x^{3}+7x^{2}\right)}.

$$\frac{\left(x-6\right)\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x\left(x-6\right)\left(2x+7\right)x^{2}}\times \frac{1}{x}$$

Cancel out x-6 in both numerator and denominator.

$$\frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x\left(2x+7\right)x^{2}}\times \frac{1}{x}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x^{3}\left(2x+7\right)}\times \frac{1}{x}$$

Multiply \frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x^{3}\left(2x+7\right)} times \frac{1}{x} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x^{3}\left(2x+7\right)x}$$

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$\frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x^{4}\left(2x+7\right)}$$

Use the distributive property to multiply x-4 by x+4 and combine like terms.

$$\frac{\left(x^{2}-16\right)\left(2x+3\right)}{4x^{4}\left(2x+7\right)}$$

Use the distributive property to multiply x^{2}-16 by 2x+3.

$$\frac{2x^{3}+3x^{2}-32x-48}{4x^{4}\left(2x+7\right)}$$

Use the distributive property to multiply 4x^{4} by 2x+7.

$$\frac{2x^{3}+3x^{2}-32x-48}{8x^{5}+28x^{4}}$$

Expand

$$\frac{2x^{3}+3x^{2}-32x-48}{4x^{2}\left(2x^{3}+7x^{2}\right)}$$

Short Solution Steps

Divide \frac{2x^{2}-9x-18}{4x^{2}-24x} by \frac{2x^{3}+7x^{2}}{x^{2}-16} by multiplying \frac{2x^{2}-9x-18}{4x^{2}-24x} by the reciprocal of \frac{2x^{3}+7x^{2}}{x^{2}-16}.

$$\frac{\left(2x^{2}-9x-18\right)\left(x^{2}-16\right)}{\left(4x^{2}-24x\right)\left(2x^{3}+7x^{2}\right)}\times \frac{1}{x}$$

Factor the expressions that are not already factored in \frac{\left(2x^{2}-9x-18\right)\left(x^{2}-16\right)}{\left(4x^{2}-24x\right)\left(2x^{3}+7x^{2}\right)}.

$$\frac{\left(x-6\right)\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x\left(x-6\right)\left(2x+7\right)x^{2}}\times \frac{1}{x}$$

Cancel out x-6 in both numerator and denominator.

$$\frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x\left(2x+7\right)x^{2}}\times \frac{1}{x}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x^{3}\left(2x+7\right)}\times \frac{1}{x}$$

Multiply \frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x^{3}\left(2x+7\right)} times \frac{1}{x} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x^{3}\left(2x+7\right)x}$$

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$\frac{\left(x-4\right)\left(x+4\right)\left(2x+3\right)}{4x^{4}\left(2x+7\right)}$$

Use the distributive property to multiply x-4 by x+4 and combine like terms.

$$\frac{\left(x^{2}-16\right)\left(2x+3\right)}{4x^{4}\left(2x+7\right)}$$

Use the distributive property to multiply x^{2}-16 by 2x+3.

$$\frac{2x^{3}+3x^{2}-32x-48}{4x^{4}\left(2x+7\right)}$$

Use the distributive property to multiply 4x^{4} by 2x+7.

$$\frac{2x^{3}+3x^{2}-32x-48}{8x^{5}+28x^{4}}$$