There can be small differences of terminology, but nothing as radical as what you are experiencing.
The function $f(x)$ has a relative maximum (also known as a local maximum) at $x=c$ if there is a positive $\epsilon$ such that $f(x)\le f(c)$ for all $x$ in the interval $(c-\epsilon, c +\epsilon)$.
Sometimes an exception is made when we are maximizing a function over a closed interval $[a,b]$. If there is an $\epsilon$ such that $f(x)\le f(a)$ for all $x$ in the interval $[a,a+\epsilon)$, then some people may say there is a relative maximum at $x=a$. I believe that most standard calculus books do not count endpoint maxima as relative maxima. Do check your text. Whatever it says is, for your purposes, the local dialect. That dialect may change a little when you get to university.
A relative maximum can be an absolute maximum. An absolute maximum can be a relative maximum. If we have an absolute maximum at $x=c$, and our function is defined in some interval $(c-\epsilon, c+\epsilon)$, then the absolute maximimum is automatically a relative maximum.
From your description, the exam is closest to using the words in the standard way. The text may also be substantially correct.
Remark: I am sorry that you are getting mixed messages. The online courses, at least in the Dogwood province, can be on the weak side.
If a polynomial is given numerically, (coefficients $a_0..a_n$ are given),
the resultant method can be used to get at a numerical value of the discriminant. The coefficients of the polynomial and its derivative are put in a (n+2)- squared Sylvester matrix. Then the determinant is the discriminant wanted. Where writing out the discriminant of a matrix containing symbols is prohibitive, the discriminant can be calculated swiftly numerically using existing matrix packages in $O(N^3)$ time.
It is very nicely explained in
http://www2.math.uu.se/~svante/papers/sjN5.pdf
This is example 4.7
If $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, then Theorem 3.3 yields
\begin{align}
\begin{array}{| ccccccc |}
& a & b & c & d & e & 0 & 0 &\\
& 0 & a & b & c & d & e & 0 \\
& 0 & 0 & a & b & c & d & e \\
& 4a& 3b& 2c& d & 0 & 0 & 0 \\
& 0 & 4a& 3b& 2c& d& 0 & 0 \\
& 0 & 0 & 4a& 3b& 2c& d & 0 \\
& 0 & 0 & 0 & 4a& 3b& 2c& d \\
\end{array}
\end{align}
= $b^2c^2d^2 - 4b^2c^3 e - 4b^3d^3 + 18b^3cde - 27b^4e^2 - 4ac^3d^2 + 16ac^4e + 18abcd^3 - 80abc^2de - 6ab^2d^2e + 144ab^2ce^2 - 27a^2d^4
+ 144a^2cd^2e - 128a^2c^2e^2 - 192a^2bde^2 + 256a^3e^3$
Best Answer
If you have a third-degree one, you can derivate your expression (this is quite easy to do by a algorithm of your own, for polynomials). Then you get the max/min by x=-b'/2a'.
For a fourth-degree polynomial, just derivate again, and then use your previous programm for 3rd-deg to find the max/min !
Once you've got an algorithm for degree n, repeat the operation for degree n+1