Solve 3x^2+4-5/x^2+5x+4-2x/x+1+4/x+4

$$\frac { 3 x ^ { 2 } + 4 - 5 } { x ^ { 2 } + 5 x + 4 } - \frac { 2 x } { x + 1 } + \frac { 4 } { x + 4 }$$

Evaluate

$$\frac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x+4\right)}$$

Short Solution Steps

Subtract 5 from 4 to get -1.

$$\frac{3x^{2}-1}{x^{2}+5x+4}-\frac{2x}{x+1}+\frac{4}{x+4}$$

Factor x^{2}+5x+4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(x+4\right) and x+1 is \left(x+1\right)\left(x+4\right). Multiply \frac{2x}{x+1} times \frac{x+4}{x+4}.

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

Since \frac{3x^{2}-1}{\left(x+1\right)\left(x+4\right)} and \frac{2x\left(x+4\right)}{\left(x+1\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3x^{2}-1-2x\left(x+4\right).

$$\frac{3x^{2}-1-2x^{2}-8x}{\left(x+1\right)\left(x+4\right)}+\frac{4}{x+4}$$

Combine like terms in 3x^{2}-1-2x^{2}-8x.

$$\frac{x^{2}-1-8x}{\left(x+1\right)\left(x+4\right)}+\frac{4}{x+4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(x+4\right) and x+4 is \left(x+1\right)\left(x+4\right). Multiply \frac{4}{x+4} times \frac{x+1}{x+1}.

$$\frac{x^{2}-1-8x}{\left(x+1\right)\left(x+4\right)}+\frac{4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}$$

Since \frac{x^{2}-1-8x}{\left(x+1\right)\left(x+4\right)} and \frac{4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}-1-8x+4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}$$

Do the multiplications in x^{2}-1-8x+4\left(x+1\right).

$$\frac{x^{2}-1-8x+4x+4}{\left(x+1\right)\left(x+4\right)}$$

Combine like terms in x^{2}-1-8x+4x+4.

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

Expand \left(x+1\right)\left(x+4\right).

$$\frac{x^{2}+3-4x}{x^{2}+5x+4}$$

Expand

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

Short Solution Steps

Subtract 5 from 4 to get -1.

$$\frac{3x^{2}-1}{x^{2}+5x+4}-\frac{2x}{x+1}+\frac{4}{x+4}$$

Factor x^{2}+5x+4.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(x+4\right) and x+1 is \left(x+1\right)\left(x+4\right). Multiply \frac{2x}{x+1} times \frac{x+4}{x+4}.

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

Since \frac{3x^{2}-1}{\left(x+1\right)\left(x+4\right)} and \frac{2x\left(x+4\right)}{\left(x+1\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3x^{2}-1-2x\left(x+4\right).

$$\frac{3x^{2}-1-2x^{2}-8x}{\left(x+1\right)\left(x+4\right)}+\frac{4}{x+4}$$

Combine like terms in 3x^{2}-1-2x^{2}-8x.

$$\frac{x^{2}-1-8x}{\left(x+1\right)\left(x+4\right)}+\frac{4}{x+4}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(x+4\right) and x+4 is \left(x+1\right)\left(x+4\right). Multiply \frac{4}{x+4} times \frac{x+1}{x+1}.

$$\frac{x^{2}-1-8x}{\left(x+1\right)\left(x+4\right)}+\frac{4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}$$

Since \frac{x^{2}-1-8x}{\left(x+1\right)\left(x+4\right)} and \frac{4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}-1-8x+4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}$$

Do the multiplications in x^{2}-1-8x+4\left(x+1\right).

$$\frac{x^{2}-1-8x+4x+4}{\left(x+1\right)\left(x+4\right)}$$

Combine like terms in x^{2}-1-8x+4x+4.

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

Expand \left(x+1\right)\left(x+4\right).

$$\frac{x^{2}+3-4x}{x^{2}+5x+4}$$