Solve 4x^4+4x^3-9x^2-x+2divx-1

$$4 x ^ { 4 } + 4 x ^ { 3 } - 9 x ^ { 2 } - x + 2 \div x - 1$$

Evaluate

$$4x^{4}+4x^{3}-9x^{2}-x-1+\frac{2}{x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 4x^{4}+4x^{3}-9x^{2}-x-1 times \frac{x}{x}.

$$\frac{\left(4x^{4}+4x^{3}-9x^{2}-x-1\right)x}{x}+\frac{2}{x}$$

Since \frac{\left(4x^{4}+4x^{3}-9x^{2}-x-1\right)x}{x} and \frac{2}{x} have the same denominator, add them by adding their numerators.

$$\frac{\left(4x^{4}+4x^{3}-9x^{2}-x-1\right)x+2}{x}$$

Do the multiplications in \left(4x^{4}+4x^{3}-9x^{2}-x-1\right)x+2.

$$\frac{4x^{5}+4x^{4}-9x^{3}-x^{2}-x+2}{x}$$

Factor

$$\frac{4x^{5}+4x^{4}-9x^{3}-x^{2}-x+2}{x}$$