(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?
 A: 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:
$$\dfrac{4x^2 + 2\cdot \color{blue}{\bf 2}(x) - 5}{x + 2} = \dfrac{4(x - 1)(x+2) + 3}{x+2},\text{ i.e.}\;\; f(x) = 4(x-1)(x+2) + 3$$
A: $4x^2+2kx-5=(x+2)2x+x(2k-4)-5$
$=(x+2)4x+(x+2)(2k-8)-2(2k-8)-5$
$4x^2+2kx-5=(x+2)(2x+2k-8)+11-4k$ 
Alternatively, putting $x=-2,4(-2)^2+2k(-2)-5=11-4k$ which we have obtained by the long division 
So, $11-4k=3\implies k=2$
So, both of the methods are correct. In fact, they are not independent.
A: 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 $11-4k = 3 \implies k = 2$.
A: It is correct, but it is not clear if you have inferred both directions, which may done as follows
$$\rm 3\, =\, (\color{#C00}{f(x)\ mod\ x\!+\!2})\iff 3\, =\, \color{#c00}{f(-2)}\, =\, 11\!-\!4k \iff k=2 $$
If what you wrote denotes only the $(\Rightarrow)$ direction above, then either you need to enhance the argument to be bidirectional, or else you need to explicitly verify that $\rm\:k = 2\:$ yields the sought remainder (as amWhy mentioned).
