A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$? Question

Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.

My attempt
Let $f(x) = 3x^3 + 7x^2 - 48x + 49$
$\text{Using Remainder Theorem,}$
\begin{align}
f(-k) &= f(k) \\
3(-k)^3  + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}
$\text{Comparing } 6k^2 + 14k - 96 = 0 \text{ with } ak^2 + bk + c = 0, a = 6, b = 14, c = -96$
\begin{align}
k = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } &= \frac{ -(14) \pm \sqrt{(14)^2 - 4(6)(-96)} }{ 2(6) } \\
k &= 3 \text{ and } -5\frac{1}{3}
\end{align}
$\therefore k = 3, -5\frac{1}{3} \text{ or } 0 $

My answer is incorrect. The correct answer is $k = 0, 4 \text{ or } -4$

By the factor theorem, $3x^3+7x^2-48x+49 = (ax+b)(x^2 - k^2) + c$. In other words, dividing by a quadratic term leaves a linear quotient and a constant remainder, for the powers on both sides to match.
Observe that there is only one way to make the terms of $x^3, x^2, x$. Thus  $ax^3 = 3x^3 \implies x= 3$ and hence $-ak^2 = -48 \implies k^2 = 16, k = ±4$.
