Solve {l}{x^9-x^5+6x^4}{-6}

$$\left. \begin{array} { l } { x ^ { 9 } - x ^ { 5 } + 6 x ^ { 4 } } \\ { - 6 } \end{array} \right.$$

Factor

$$\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{5}+6\right)$$

Solution Steps

Do the grouping x^{9}-x^{5}+6x^{4}-6=\left(x^{9}-x^{5}\right)+\left(6x^{4}-6\right), and factor out x^{5} in the first and 6 in the second group.

$$x^{5}\left(x^{4}-1\right)+6\left(x^{4}-1\right)$$

Factor out common term x^{4}-1 by using distributive property.

$$\left(x^{4}-1\right)\left(x^{5}+6\right)$$

Consider x^{4}-1. Rewrite x^{4}-1 as \left(x^{2}\right)^{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).

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

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. The following polynomials are not factored since they do not have any rational roots: x^{2}+1,x^{5}+6.

$$\left(x-1\right)\left(x+1\right)\left(x^{2}+1\right)\left(x^{5}+6\right)$$

Evaluate

$$x^{9}-x^{5}+6x^{4}-6$$