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!
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.
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.