The remainder when $x^2$ $+ ax +4$ is divided by $x-4$ is four times the remainder when the same expression is divided by $x-1$. Find the value of $a$ According to my textbook, the answer to this is $-2$.
Isn't this true for any value of a though?
I started by doing the following:

$4a+20=4x$
  $a+5  = x$

I'm not sure how they got $-2$ from this, as both equations are equivalent.

(This is not an answer, but it is too long for a comment.)
Let $f(x) = x^2 + ax + 4$.
I can't make any sense of what you claim to get on dividing $f(x)$ by $x-4$, "$(x-4)(x+4+a)(4a+20)$" or on dividing $f(x)$ by $x-1$, "$(x-1)(x+1+a)(a+5)$".  You also claim the first is the second multiplied by four, which is evidently false.  Why?    The expressions you write must contain "$a^2$"s, but $f(x)$ does not.  \begin{align*}
(x-4)&(x+4+a)(4a+20) \\ &= 4 a x^2 + 20 x^2 + 4 a^2 x + 20 a x - 16 a^2 - 144 a - 320  \\
(x-1)&(x+1+a)(a+5)  \\ &= a x^2 + 5 x^2 + a^2 x + 5 a x - a^2 - 6 a - 5
\end{align*}
It is also straightforward to see that the first is not the second multiplied by four.  In particular, neither $x+4+a = 4(x+1+a)$ nor $x+4+a = x+1+a$.
