Universal factoring method or list of methods (trinomials) I am a student in calculus II. I'm now failing tests solely because I cannot factor; I understand everything else. This is compounded by the fact it seems to exceedingly hard to find anything comprehensive online. This is a long-running problem I've been facing; I can't be the only one.
I know some methods, but I always run into more problems I can't factor. I know about synthetic/long division, binomial/perfect square trinomial factoring, difference and sum of cubes and squares, the grouping method, the quadratic formula (unsure how to use for factoring), guess and check, and some trinomial methods. I only know the difference/sum of squares/cubes well. For example, I can't factor $4x^2 - 4x - 3$.
It would be dandy if I could get a comprehensive list or a universal solution to all these factoring questions, but that's unlikely to happen (and would be too broad). I'll narrow this down to something more tangible: What are the methods I should know to factor trinomials?

Before we consider how to factor $4x^2 - 4x - 3$, let's see what happens when we form the product of two linear monomials.
\begin{align*}
(6x - 7)(3x + 5) & = 6x(3x + 5) - 7(3x + 5)\\
                 & = \color{blue}{18}x^2 + \color{green}{30}x \color{green}{ - 21}{x} \color{blue}{- 35}\\
                 & = \color{blue}{18}x^2 - \color{green}{9}x - \color{blue}{35}
\end{align*}
To factor, what we wish to do is carry out the steps in reverse.  The key observation is that in the expression $\color{blue}{18}x^2 + \color{green}{30}x \color{green}{ - 21}{x} \color{blue}{- 35}$, the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients whose sum is the linear coefficient of the product.  That is, 
$$(\color{blue}{18})(\color{blue}{-35}) = (\color{green}{30})(\color{green}{-21}) = -630$$
This suggests that we can factor a quadratic polynomial $ax^2 + bx + c$ with rational coefficients with respect to the rationals if we can split the linear term if we can find two numbers with product $ac$ and sum $b$.
In your example of $4x^2 - 4x - 3$, we can split the linear term if we can find two numbers with product $(\color{blue}{4})(\color{blue}{-3}) = -12$ and sum $-4$.  The possible factorizations of $-12$ are
\begin{align*}
-12 & = -1 \cdot 12 & -12 & = 1 \cdot -12\\
    & = -2 \cdot 6 & & = \color{green}{2} \cdot \color{green}{-6}\\
    & = -3 \cdot 4 & & = 3 \cdot -4
\end{align*}
By inspection, the only pair of factors of $-12$ with sum $-4$ are $2$ and $-6$.  Hence,
\begin{align*}
4x^2 - 4x - 3 & = 4x^2 + 2x - 6x - 3 && \text{split the linear term}\\
              & = 2x(2x + 1) - 3(2x + 1) && \text{factor by grouping}\\
              & = (2x - 3)(2x + 1) && \text{extract the common factor}
\end{align*}
Another option is completing the square.
\begin{align*}
4x^2 - 4x - 3 & = 4(x^2 - x) - 3\\
              & = 4\left(x^2 - x + \frac{1}{4}\right) - 1 - 3\\
              & = 4\left(x - \frac{1}{2}\right)^2 - 4\\
              & = 4\left[\left(x - \frac{1}{2}\right)^2 - 1\right]\\
              & = 4\left[\left(x - \frac{1}{2}\right) + 1\right]\left[\left(x - \frac{1}{2}\right) - 1\right]\\
              & = 4\left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right)
\end{align*}
Since 
\begin{align*}
4\left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right) & = 2 \cdot 2 \left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right)\\ 
& = \left[2\left(x + \frac{1}{2}\right)\right]\left[2\left(x - \frac{3}{2}\right)\right]\\
& = (2x + 1)(2x - 3)
\end{align*}
our two factorizations are equivalent.
For the general case,
\begin{align*}
ax^2 + bx + c & = a\left(x^2 + \frac{b}{a}x\right) + c\\
              & = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) - \frac{b^2}{4a} + c\\
              & = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}\\
              & = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right]\\
              & = a\left[x + \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}\right]\left[x + \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}\right]
\end{align*}
