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 $a$

Question

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.

Answer 1

Let $$f(x)= x^2 + ax +4$$

You want $$f(4) = 4f(1)$$ $$ f(4) = 20+4a$$

and $$f(1) =5+a$$

Thus $$20+4a=4(5+a)$$ which is valid for every $a$

Therefore $a$ could be any number that we like.

Answer 2

(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 $x+4+a = x+1+a$.