Given that p(x) leaves a remainder of R when divided by (x-1), we have p(1) = R.

Given that $p(x)$ leaves a remainder of $R$ when divided by $(x-1)$, we have $p(1) = R$.
Substitute $x=1$ into $p(x) = x^3 + kx^2 - x + 12$:

Given that $p(x)$ leaves a remainder of $8R$ when divided by $(x-4)$, we have $p(4) = 8R$.
Substitute $x=4$ into $p(x) = x^3 + kx^2 - x + 12$:

Substitute equation $(*)$ into equation $(**)$:

Now substitute $k=3$ back into equation $(*)$ to find $R$:

The values are $\boxed{R=15, k=3}$.

Using $k=3$, the polynomial is $p(x) = x^3 + 3x^2 - x + 12$.

For $(x+4)$ to be a factor of $p(x)$, we must show that $p(-4) = 0$.
Substitute $x=-4$ into $p(x)$:

Since $p(-4) = 0$, by the Factor Theorem, $(x+4)$ is a factor of $p(x)$.