Solve 2x^4+x^3-6x^2+1divx+2

$$2 x ^ { 4 } + x ^ { 3 } - 6 x ^ { 2 } + 1 \div x + 2$$

Evaluate

$$2x^{4}+x^{3}-6x^{2}+2+\frac{1}{x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x^{4}+x^{3}-6x^{2}+2 times \frac{x}{x}.

$$\frac{\left(2x^{4}+x^{3}-6x^{2}+2\right)x}{x}+\frac{1}{x}$$

Since \frac{\left(2x^{4}+x^{3}-6x^{2}+2\right)x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

$$\frac{\left(2x^{4}+x^{3}-6x^{2}+2\right)x+1}{x}$$

Do the multiplications in \left(2x^{4}+x^{3}-6x^{2}+2\right)x+1.

$$\frac{2x^{5}+x^{4}-6x^{3}+2x+1}{x}$$

Factor

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