(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?

We have
\begin{align}
\dfrac{4x^2 + 2kx - 5}{x+2} & = \dfrac{4(x+2)^2 - 4 \cdot(4x) - 4 \cdot 4 + 2kx-5}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)x - 21}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)x - 21}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)(x+2) - 21-2(2k-16)}{x+2}\\
& = 4(x+2) + (2k-16) + \dfrac{11-4k}{x+2}
\end{align}
This means $11-4k = 3 \implies k = 2$.
