Find the value of $k$ such that $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$. $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$.
(a) -1 (b) 1 (c) 2 (d) -2
I am getting the value of k: $-17/29$ after equating the remainders.
$p(x)= kx^3 + 4x^2 + 3x - 4/(x – 3)$: remainder= $(30k+32)$ 
$q(x)= x^3 - 4x + k/(x – 3)$: remainder= $(k+15)$
So, 
$30k+32=k+15$
=> $k=(-17/29)$
Any help would be appreciated. :)

By the polynomial remainder theorem, the remainder of $f(x)$ when divided by $x - a$ is equal to $f(a)$. So we have
$$p(3) = q(3)$$
$$27k + 41 = 15 + k$$
$$k = -1$$
