Solve x^3-7x^2/x+4div(3x/x^2-16-x/x+4)

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

Evaluate

$$-x\left(x-4\right)$$

Short Solution Steps

Factor x^{2}-16.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-4\right)\left(x+4\right) and x+4 is \left(x-4\right)\left(x+4\right). Multiply \frac{x}{x+4} times \frac{x-4}{x-4}.

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

Since \frac{3x}{\left(x-4\right)\left(x+4\right)} and \frac{x\left(x-4\right)}{\left(x-4\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3x-x\left(x-4\right).

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

Combine like terms in 3x-x^{2}+4x.

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

Divide \frac{x^{3}-7x^{2}}{x+4} by \frac{7x-x^{2}}{\left(x-4\right)\left(x+4\right)} by multiplying \frac{x^{3}-7x^{2}}{x+4} by the reciprocal of \frac{7x-x^{2}}{\left(x-4\right)\left(x+4\right)}.

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

Cancel out x+4 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in -7+x.

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

Cancel out x\left(-x+7\right) in both numerator and denominator.

$$-x\left(x-4\right)$$

Expand the expression.

$$-x^{2}+4x$$

Expand

$$4x-x^{2}$$

Short Solution Steps

Factor x^{2}-16.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-4\right)\left(x+4\right) and x+4 is \left(x-4\right)\left(x+4\right). Multiply \frac{x}{x+4} times \frac{x-4}{x-4}.

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

Since \frac{3x}{\left(x-4\right)\left(x+4\right)} and \frac{x\left(x-4\right)}{\left(x-4\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3x-x\left(x-4\right).

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

Combine like terms in 3x-x^{2}+4x.

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

Divide \frac{x^{3}-7x^{2}}{x+4} by \frac{7x-x^{2}}{\left(x-4\right)\left(x+4\right)} by multiplying \frac{x^{3}-7x^{2}}{x+4} by the reciprocal of \frac{7x-x^{2}}{\left(x-4\right)\left(x+4\right)}.

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

Cancel out x+4 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in -7+x.

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

Cancel out x\left(-x+7\right) in both numerator and denominator.

$$-x\left(x-4\right)$$

Expand the expression.

$$-x^{2}+4x$$