Answer
false.
the complete factors are
$(x^2)^2-4^2 =(x^2+4)(x+2)(x-2)$.
Work Step by Step
An expression is factored completely if it is written as a product of factors so that none of its factors can be further factored.
Rewrite the given expression as a difference of squares.
$x^4-16=(x^2)^2-4^2$
Use the algebraic identity.
$a^2-b^2=(a+b)(a-b)$
We have $a=x^2$ and $b=4$.
$=(x^2)^2-4^2 =(x^2+4)(x^2-4) $
$=(x^2)^2-4^2 =(x^2+4)(x^2-2^2) $
Again use algebraic identity.
$=(x^2)^2-4^2 =(x^2+4)(x+2)(x-2)$.
The given statement is false because the factor $x^2-4$ can be further factored.
