<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] $4x^2+2kx-5=(x+2)2x+x(2k-4)-5$
$=(x+2)4x+(x+2)(2k-8)-2(2k-8)-5$
$4x^2+2kx-5=(x+2)(2x+2k-8)+11-4k$ 
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.
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