Solve x^5-2x^4-x^3+2x^2+0

$$x ^ { 5 } - 2 x ^ { 4 } - x ^ { 3 } + 2 x ^ { 2 } + 0$$

Factor

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

Solution Steps

Factor out x^{2}.

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

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

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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

Solution Steps

Anything plus zero gives itself.

$$x^{5}-2x^{4}-x^{3}+2x^{2}$$