Find all integers n such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients. I am unsure of how to approach this problem. I have thought about using the Rational root theorem, but I am unsure if this answers the question being asked.
Using the theorem, I get $\frac{p}{q} = \pm 1, \pm 13, \pm \frac{1}{5}$, and $\pm \frac{13}{5}$ as possible roots. Then I use synthetic division and Horner's method to get a remainder of $-(n+8)$. For this to be a solution, $-(n+8)=0$, so $n = -8$. Then I could do this for $+1, +13, -13,$ etc.
Is this the correct approach to answering the original question? Original question: 

Find all integers $n$ such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients.

Why would I need to have a rational root to answer the problem? Couldn't I have complex solutions where I can express $5x^2 + nx - 13$ (where n is an integer) as a product of two linear factors with integer coefficients?
I greatly appreciate any insight you could provide on this. It's been about 2 years since I've done any mathematics (a brief foray into Neuroscience turned into a longer expedition than intended) and I am longing to return to the beautiful realm of mathematics. Thanks for your time in reading through this jumbled mathematical thought!
 A: More simply, note that $5$ is prime, so the linear factors have to have coefficients of $x$ as $1$ and $5$ (or $-1$ and $-5$). 
So, you have $(x-r)(5x-s) = 5 x^2 -nx -13$ for integers $r,s$. FOILING the left hand side gives $5x^2 - (5r+s) x + rs$, so $rs=-13$ and $5r+s=n$. Since $13$ is prime, we must have $r=-1,s=13$ or $r=1,s=-13$ to meet the constraint that $rs=-13$. You can then plug these in and find $n$.
You can do the same thing with $(-x -r) (-5x-s) = 5 x^2 -nx -13$ to get the other case. 
A: We want to find $n$ such that $f(x) = 5x^2 +nx -13$ is a product of two linear terms with integer coefficients. Now let $f(x)=(ax+b)(cx+d) = acx^2 +(ad+bc)x +bd $ with $a,b,c,d$ integer. Then 
$
ac= 5 , bd =-13 , ad+bc =n$. So $(a,c)=(5,1 ),(-5,-1)$ and $(b,d)= (13,-1),(-13,1)$. This implies $n = ad+bc$ can be $\pm 64, \pm 8$
A: This can be done if and only if $5x^2+nx-13$ is a difference of two squares. $$5x^2+nx-13=\frac15(25x^2+5nx+\left(\frac{n}{2}\right)^2)-(13+\frac{n^2}{4})$$ $$5x^2+nx-13=\frac15(5x+\frac{n}{2})^2-\frac14(n^2+52).$$ Therefore, there should be an integer $m$ such that $$n^2+52=5m^2.$$ I think this will helps you. Continue from here.
A: If a quadratic polynomial $ax^2 + bx + c$ with integer coefficients, $a \neq 0$, has factorization 
$$ax^2 + bx + c = (rx + s)(tx + u) \tag{1}$$
then 
\begin{align*}
a & = rt\\
b & = ru + st\\
c & = su
\end{align*}
which can be demonstrated by substituting $0$, $1$, and $-1$ in equation 1, then solving the resulting system of equations.  Note that $ac = (rt)(su) = rstu = (ru)(st)$, that is, $b$ is the sum of two integers whose product is $ac$.
For the equation $5x^2 + nx - 13$, this means that $n$ is the sum of two integers with product $5 \cdot -13 = -65$.  Since 
\begin{align*}
-65 & = 1 \cdot -65 & -65 & = -1 \cdot 65\\
    & = 5 \cdot -13 & & = -5 \cdot 13
\end{align*}
the possible integer values for $n$ are 
\begin{align*}
1 + (-65) & = -64 & -1 + 65 & = 64\\
5 + (-13) & = -8 & -5 + 13 & = 8
\end{align*}
A: As mentioned in the comments, the discriminant $\sqrt{n^2-4 \cdot 5 \cdot -13} = \sqrt{n^2+260}$ needs to be an integer (such as $k$) for both factors to have integer coefficients.
Let $n^2+260 = k^2$ where $k$ is an integer. Then $260 = k^2-n^2 = (k+n)(k-n)$, and factorising $260 = 2^2 \times 5 \times 13$, the factor pairs of $260$ are $(1, 260); (2, 130); (4, 65); (5, 52); (10, 26); (13, 20)$.
Discard all pairs where the numbers have opposite parity: (odd, even) or (even, odd), which leaves only $(2,130)$ and $(10,26)$. As to why this is, what happens to $k$ and $n$, which must be integers?
Therefore:
$$k + n = 130, k - n = 2 \Rightarrow n = 64 \tag{1}$$
$$k + n = 2, k - n = 130 \Rightarrow n = -64 \tag{2}$$
$$k+n=26, k-n=10 \Rightarrow n=8 \tag{3}$$
$$k+n=10, k-n=26 \Rightarrow n=-8 \tag{4}$$
which is no surprise, as substituting $n \to -n$ into $(1)$ gives $(2)$, and the same into $(3)$ gives $(4)$, and vice versa.
