Answer
False. The second factor $(x^{2}-4)$ is itself a difference of two squares and can be factored.
$(x^{4}-16)= (x^{2}+4)(x+2)(x-2)$
Work Step by Step
$(x^{4}-16) = (x^{2}+4)(x^{2}-4)$
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.
