Remainder = $1$ $(x+4)$ is not a factor of $f(x)$.
Work Step by Step
The Remainder Theorem states that when a function $f(x)$ is divided by $(x-R)$ , then the remainder will be: $f(R)$. We have: $f(x)=4x^6-64x^4+x^2-15$ $\implies f(-4)=4(-4)^6-64(-4)^4+(-4)^2-15=1 \ne 0$ The Factor Theorem states that if $f(a)=0$, then $(x-a)$ is a factor of $f(x)$ and vice versa. Therefore, by the Factor Theorem $(x+4)$ is not a factor of $f(x)$.