## Precalculus (6th Edition) Blitzer

False. The correct statement is: $x^4-16$ is factored completely as $(x^2+4)(x-2)(x+2)$.
The statement is false. $x^2-4=x^2-2^2$, a difference of two squares. A difference of two squares can still be factored using the formula: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares using the formula above with $a=x$ and $b=2$ to obtain: $x^2-2^2=(x-2)(x+2)$ Thus, the completely factored form of $x^4-16$ is: $=(x^2+4)(x-2)(x+2)$