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!
Best Answer
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.