Solve x^2+2x-3quad4/x+3quad1/x-1

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

Evaluate

$$\frac{x^{4}+4x^{3}+x^{2}-6x-12}{\left(x-1\right)\left(x+3\right)}$$

Short Solution Steps

Express 3\times \frac{4}{x+3} as a single fraction.

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

Multiply \frac{3\times 4}{x+3} times \frac{1}{x-1} by multiplying numerator times numerator and denominator times denominator.

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

Multiply 3 and 4 to get 12.

$$x^{2}+2x-\frac{12}{\left(x+3\right)\left(x-1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}+2x times \frac{\left(x+3\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}.

$$\frac{\left(x^{2}+2x\right)\left(x+3\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{12}{\left(x+3\right)\left(x-1\right)}$$

Since \frac{\left(x^{2}+2x\right)\left(x+3\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)} and \frac{12}{\left(x+3\right)\left(x-1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x^{2}+2x\right)\left(x+3\right)\left(x-1\right)-12}{\left(x+3\right)\left(x-1\right)}$$

Do the multiplications in \left(x^{2}+2x\right)\left(x+3\right)\left(x-1\right)-12.

$$\frac{x^{4}+2x^{3}-3x^{2}+2x^{3}+4x^{2}-6x-12}{\left(x+3\right)\left(x-1\right)}$$

Combine like terms in x^{4}+2x^{3}-3x^{2}+2x^{3}+4x^{2}-6x-12.

$$\frac{x^{4}+4x^{3}+x^{2}-6x-12}{\left(x+3\right)\left(x-1\right)}$$

Expand \left(x+3\right)\left(x-1\right).

$$\frac{x^{4}+4x^{3}+x^{2}-6x-12}{x^{2}+2x-3}$$

Differentiate w.r.t. x

$$\frac{2\left(x+1\right)\left(x^{4}+4x^{3}-2x^{2}-12x+21\right)}{\left(\left(x-1\right)\left(x+3\right)\right)^{2}}$$