How do you factorize quadratics when the coefficient of $x^2 \gt 1$? So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?

We can factor $2x^2 - x - 3$ with respect to the rationals if we can find two numbers with product $2 \cdot -3 = -6$ and sum $-1$.  The factors of $-6$ are 
\begin{align*}
-6 & = 1 \cdot -6 & -6 & = -1 \cdot 6\\
   & = \color{blue}{2 \cdot -3} & & = -2 \cdot 3
\end{align*}
Of these four pairs of factors, only $2$ and $-3$ have sum $-1$.  Hence,
\begin{align*}
2x^2 - x - 3 & = 2x^2 + 2x - 3x - 3 && \text{split the linear term}\\
             & = 2x(x + 1) - 3(x + 1) && \text{factor by grouping}\\
             & = (2x - 3)(x + 1) && \text{extract the common factor}
\end{align*}
Note that if we multiply $2x - 3$ and $x + 1$, we carry out the steps of the factorization in reverse order.
Why does this work?  Suppose we have the factorization with respect to the rationals 
\begin{align*}
ax^2 + bx + c & = (rx + s)(tx + u)\\
              & = rx(tx + u) + s(tx + u)\\
              & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\
              & = \color{blue}{rt}x^2 + (\color{green}{ru + st})x + \color{blue}{su}
\end{align*}
Observe that the product of the coefficients of the quadratic and constant terms is equal to the product of the two coefficients that sum to the coefficient of the linear term, that is
$$(\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st}) = rstu$$
Matching coefficients gives $a = rt$, $b = ru + st$, and $c = su$.  Thus, we can factor a quadratic with respect to the rationals if we can find two numbers with product $ac$ and sum $b$.
