Solve 3:x^3-4x^2-9x+36

$$3 : x ^ { 3 } - 4 x ^ { 2 } - 9 x + 36$$

Evaluate

$$-4x^{2}-9x+36+\frac{3}{x^{3}}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -4x^{2}-9x+36 times \frac{x^{3}}{x^{3}}.

$$\frac{3}{x^{3}}+\frac{\left(-4x^{2}-9x+36\right)x^{3}}{x^{3}}$$

Since \frac{3}{x^{3}} and \frac{\left(-4x^{2}-9x+36\right)x^{3}}{x^{3}} have the same denominator, add them by adding their numerators.

$$\frac{3+\left(-4x^{2}-9x+36\right)x^{3}}{x^{3}}$$

Do the multiplications in 3+\left(-4x^{2}-9x+36\right)x^{3}.

$$\frac{3-4x^{5}-9x^{4}+36x^{3}}{x^{3}}$$

Factor

$$-\frac{4x^{5}+9x^{4}-36x^{3}-3}{x^{3}}$$