Solve x-1divx^3-7x^2+19x-13

$$x-1 \div { x }^{ 3 } -7 { x }^{ 2 } +19x-13$$

Evaluate

$$-7x^{2}+20x-13-\frac{1}{x^{3}}$$

Short Solution Steps

Combine x and 19x to get 20x.

$$20x-\frac{1}{x^{3}}-7x^{2}-13$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 20x-7x^{2}-13 times \frac{x^{3}}{x^{3}}.

$$\frac{\left(20x-7x^{2}-13\right)x^{3}}{x^{3}}-\frac{1}{x^{3}}$$

Since \frac{\left(20x-7x^{2}-13\right)x^{3}}{x^{3}} and \frac{1}{x^{3}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(20x-7x^{2}-13\right)x^{3}-1}{x^{3}}$$

Do the multiplications in \left(20x-7x^{2}-13\right)x^{3}-1.

$$\frac{20x^{4}-7x^{5}-13x^{3}-1}{x^{3}}$$

Factor

$$-\frac{7x^{5}-20x^{4}+13x^{3}+1}{x^{3}}$$