<s>[INST] (4x^2+2kx-5)/(x+2) remainder is 3 find value of k? 2 methods - first is long division by $(x+2)$, 2nd is to use remainder theorem
let $f(x) = 4x^2+2kx-5$ and $g(x) = x+2$
to find the remainder of $\frac{f(x)}{g(x)}$ where $g(x) = (x+c)$ we need to evaluate $f(-c)$
$f(-2) = 4(-2)^2+2k(-2) -5$
Because the remainder is 3, we know that $f(-2)=3$
so $$16 - 4k -5 = 3$$
$$4k = 8$$
$$k=2$$
is that correct?
 [/INST] Your method works well here. And indeed, $k = 2$.  It's always a good idea to "check out" whether the equation, with $k = 2$, divided by $x + 2$, gives a remainder of $3$.
Substituting $\color{blue}{\bf k = 2}$ into $\dfrac{f(x)}{g(x)}$ gives us:
$$\dfrac{4x^2 + 2\cdot \color{blue}{\bf 2}(x) - 5}{x + 2} = \dfrac{4(x - 1)(x+2) + 3}{x+2},\text{ i.e.}\;\; f(x) = 4(x-1)(x+2) + 3$$
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