## Precalculus (6th Edition) Blitzer

$7(x^2+1)(x+1)(x-1)$
Factor out the GCF of $7$ to obtain: $=7(x^4-1)$ The binomial above can be written as $(x^2)^2-1^2$, so the expression above is equivalent to: $=7[(x^2)^2-1^2]$ Factor the difference of two squares using the formula $a^2-b^2=(a+b)(a-b)$ to obtain: $=7(x^2+1)(x^2-1)$ Factor the difference of two squares using the same formula to obtain: $=7(x^2+1)(x+1)(x-1)$