Finding the values of $k$ for which $x + 2$ is a factor of $(x + 1)^5 + (5x + k)^3$ $$ \huge\text{Question} $$

Find the values of $k$ for which $x + 2$ is a factor of $(x + 1)^5 + (5x + k)^3$.


$$ \huge\text{My attempt} $$
$$ \text{Let } f(x) = (x + 1)^5 + (5x + k)^3 $$

$$ \text{Since } x + 2 \text{ is a factor of } f(x), f(-2) = 0, \text{ i.e.} $$
\begin{align}
[(-2) + 1]^5 + [5(-2) + k]^3 &amp;= 0 \\
(-2 + 1)^5 + (-10 + k)^3 &amp;= 0 \\
(-1)^5 + (-10 + k)(-10 + k)(-10 + k) &amp;= 0 \\
-1 + (-10 + k)[-10(-10 + k) + k(-10 + k)] &amp;= 0 \\
-1 + (-10 + k)(100 - 10k - 10k + k^2) &amp;= 0 \\
-1 + (-10 + k)(100 - 20k + k^2) &amp;= 0 \\
-1 + k(100 - 20k + k^2) -10(100 - 20k + k^2) &amp;= 0 \\
-1 + 100k - 20k^2 + k^3 - 1000 + 200k - 10k^2 &amp;= 0 \\
k^3 - 20k^2 - 10k^2 + 200k + 100k - 1000 - 1 &amp;= 0 \\
k^3 - 30k^2 + 300k - 1001 &amp;= 0
\end{align}

$$ \text{Was my approach correct? How should I continue solving the problem?} $$
You should not expand the power : as you said, you want to find $k$ such that
$$(k-10)^3=1$$
which is equivalent to
$$k-10=1$$
so
$$k=11$$
Your first insight is exactly the idea, though the algebra would be easier without the expansion you do:
$$f(-2) = 0 \implies -1 + (k - 10)^3 = 0 \implies (k - 10)^3 = 1$$
If your problem only asks for real solutions of $k$, then it is clear that $k = 11$ is the only solution. However, if complex numbers are allowed, you will need to consider the case where $k - 10$ is a complex third root of unity, i.e. $k - 10 = \frac{-1}{ 2} + \frac{\sqrt 3}{ 2} i$ or $k - 10 = \frac{-1}{ 2} - \frac{\sqrt 3}{ 2} i$.