Solve 16q^9-16q

$$16 q ^ { 9 } - 16 q$$

Factor

$$16q\left(q-1\right)\left(q+1\right)\left(q^{2}+1\right)\left(q^{4}+1\right)$$

Solution Steps

Factor out 16.

$$16\left(q^{9}-q\right)$$

Consider q^{9}-q. Factor out q.

$$q\left(q^{8}-1\right)$$

Consider q^{8}-1. Rewrite q^{8}-1 as \left(q^{4}\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).

$$\left(q^{4}-1\right)\left(q^{4}+1\right)$$

Consider q^{4}-1. Rewrite q^{4}-1 as \left(q^{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).

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

Consider q^{2}-1. Rewrite q^{2}-1 as q^{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).

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: q^{2}+1,q^{4}+1.

$$16q\left(q-1\right)\left(q+1\right)\left(q^{2}+1\right)\left(q^{4}+1\right)$$

Evaluate

$$16q\left(q^{8}-1\right)$$