## College Algebra (6th Edition)

False. The second factor $(x^{2}-4)$ is itself a difference of two squares and can be factored. $(x^{4}-16)= (x^{2}+4)(x+2)(x-2)$
$(x^{4}-16) = (x^{2}+4)(x^{2}-4)$ The second factor $(x^{2}-4)$ is itself a difference of two squares and can be factored further. Express $(x^{2}-4)$ as the difference of two squares $x^{2}-2^{2}$ $(x^{4}-16) = (x^{2}+4)(x^{2}-2^{2})$ Using the formula $[(a^{2}-b^{2}) = (a+b)(a-b)]$ $(x^{4}-16)= (x^{2}+4)(x+2)(x-2)$ factored completely.