Solve x^6+2x^5-2x^3-x^2

$$x ^ { 6 } + 2 x ^ { 5 } - 2 x ^ { 3 } - x ^ { 2 }$$

Factor

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

Solution Steps

Factor out x^{2}.

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

Consider x^{4}+2x^{3}-2x-1. Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{4} and m divides the constant factor -1. One such factor is x^{2}-1. Factor the polynomial by dividing it by this factor.

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

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).

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

Consider x^{2}+2x+1. Use the perfect square formula, a^{2}+2ab+b^{2}=\left(a+b\right)^{2}, where a=x and b=1.

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

Rewrite the complete factored expression.

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

Evaluate

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