Solve 1-x^2+x^3-x^5

$$1 - x ^ { 2 } + x ^ { 3 } - x ^ { 5 }$$

Factor

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

Solution Steps

Do the grouping 1-x^{2}+x^{3}-x^{5}=\left(1-x^{2}\right)+\left(x^{3}-x^{5}\right), and factor out x^{3} in x^{3}-x^{5}.

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

Factor out common term -x^{2}+1 by using distributive property.

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

Consider -x^{2}+1. Rewrite -x^{2}+1 as 1^{2}-x^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(1-x\right)\left(1+x\right)$$

Reorder the terms.

$$\left(-x+1\right)\left(x+1\right)$$

Consider x^{3}+1. Rewrite x^{3}+1 as x^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Rewrite the complete factored expression. Polynomial x^{2}-x+1 is not factored since it does not have any rational roots.

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

Evaluate

$$1-x^{2}+x^{3}-x^{5}$$