Solve 3x^2+2x-8*1/4x-12

$$3 x ^ { 2 } + 2 x - 8 \cdot \frac { 1 } { 4 x - 12 }$$

Evaluate

$$\frac{3x^{3}-7x^{2}-6x-2}{x-3}$$

Short Solution Steps

Express 8\times \frac{1}{4x-12} as a single fraction.

$$3x^{2}+2x-\frac{8}{4x-12}$$

Factor 4x-12.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3x^{2}+2x times \frac{4\left(x-3\right)}{4\left(x-3\right)}.

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

Since \frac{\left(3x^{2}+2x\right)\times 4\left(x-3\right)}{4\left(x-3\right)} and \frac{8}{4\left(x-3\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(3x^{2}+2x\right)\times 4\left(x-3\right)-8.

$$\frac{12x^{3}-36x^{2}+8x^{2}-24x-8}{4\left(x-3\right)}$$

Combine like terms in 12x^{3}-36x^{2}+8x^{2}-24x-8.

$$\frac{12x^{3}-28x^{2}-24x-8}{4\left(x-3\right)}$$

Factor the expressions that are not already factored in \frac{12x^{3}-28x^{2}-24x-8}{4\left(x-3\right)}.

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

Cancel out 4 in both numerator and denominator.

$$\frac{3x^{3}-7x^{2}-6x-2}{x-3}$$

Differentiate w.r.t. x

$$\frac{2\left(3x^{3}-17x^{2}+21x+10\right)}{\left(x-3\right)^{2}}$$