Solve =x-1/2x^2+1/6x^3-1/12x^4+

$$= x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 6 } x ^ { 3 } - \frac { 1 } { 12 } x ^ { 4 } +$$

Factor

$$\frac{x\left(2-x\right)\left(x^{2}+6\right)}{12}$$

Solution Steps

Factor out \frac{1}{12}.

$$\frac{12x-6x^{2}+2x^{3}-x^{4}}{12}$$

Consider 12x-6x^{2}+2x^{3}-x^{4}. Factor out x.

$$x\left(12-6x+2x^{2}-x^{3}\right)$$

Consider 12-6x+2x^{2}-x^{3}. Do the grouping 12-6x+2x^{2}-x^{3}=\left(12-6x\right)+\left(2x^{2}-x^{3}\right), and factor out 6 in the first and x^{2} in the second group.

$$6\left(-x+2\right)+x^{2}\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(6+x^{2}\right)$$

Rewrite the complete factored expression. Polynomial x^{2}+6 is not factored since it does not have any rational roots.

$$\frac{x\left(-x+2\right)\left(x^{2}+6\right)}{12}$$

Evaluate

$$\frac{x\left(2-x\right)\left(x^{2}+6\right)}{12}$$