Solve 2x^3-3x^2-3x+7divx-1

$$2 { x }^{ 3 } -3 { x }^{ 2 } -3x+7 \div x-1$$

Evaluate

$$2x^{3}-3x^{2}-3x-1+\frac{7}{x}$$

Short Solution Steps

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

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

Since \frac{\left(2x^{3}-3x^{2}-3x-1\right)x}{x} and \frac{7}{x} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(2x^{3}-3x^{2}-3x-1\right)x+7.

$$\frac{2x^{4}-3x^{3}-3x^{2}-x+7}{x}$$

Factor

$$\frac{2x^{4}-3x^{3}-3x^{2}-x+7}{x}$$