Solve x^3-8x^2+19x-12divx-4

$${ x }^{ 3 } -8 { x }^{ 2 } +19x-12 \div x-4$$

Evaluate

$$x^{3}-8x^{2}+19x-4-\frac{12}{x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{3}-8x^{2}+19x-4 times \frac{x}{x}.

$$\frac{\left(x^{3}-8x^{2}+19x-4\right)x}{x}-\frac{12}{x}$$

Since \frac{\left(x^{3}-8x^{2}+19x-4\right)x}{x} and \frac{12}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(x^{3}-8x^{2}+19x-4\right)x-12}{x}$$

Do the multiplications in \left(x^{3}-8x^{2}+19x-4\right)x-12.

$$\frac{x^{4}-8x^{3}+19x^{2}-4x-12}{x}$$

Factor

$$\frac{x^{4}-8x^{3}+19x^{2}-4x-12}{x}$$