Taking Calculus in a few days and I still don't know how to factorize quadratics Taking Calculus in a few days and I still don't know how to factorize quadratics with a coefficient in front of the 'x' term. I just don't understand any explanation. My teacher gave up and said just use the formula to find the roots or something like that..
Can someone explain to me simply how I would step by step factorize something like $4x^2 + 16x - 19$ ?

The term $4x^2+16x$ is almost square of $2x+4$, more precisely
$$
4x^2+16x=(2x+4)^2-4^2.
$$
Therefore, we have
\begin{align}
4x^2+16x-19&=(2x+4)^2-4^2-19\\
&=(2x+4)^2-35\\
&=(2x+4)^2-\left(\sqrt{35}\right)^2.
\end{align}
Knowing that
$$
p^2-q^2=(p-q)(p+q),\tag1
$$
then we have
\begin{align}
4x^2+16x-19
&=\left(2x+4-\sqrt{35}\ \right)\left(2x+4+\sqrt{35}\ \right).\qquad\blacksquare
\end{align}

Addendum :
Here is a general approach to factorize a quadratic equation. Suppose that we want to factorize quadratic equation
$$
ax^2+bx+c=0.\tag2
$$
Now, multiplying $(2)$ by $4a$ yields
$$
4a^2x^2+4abx+4ac=0.\tag3
$$
The term $4a^2x^2+4abx$ is almost square of $2ax+b$, more precisely
$$
4a^2x^2+4abx=(2ax+b)^2-b^2,\tag4
$$
then $(3)$ turns out to be
\begin{align}
4a^2x^2+4abx+4ac&=(2ax+b)^2-b^2+4ac\\
&=(2ax+b)^2-(b^2-4ac)\\
&=(2ax+b)^2-\left(\sqrt{b^2-4ac}\right)^2.
\end{align}
Using $(1)$, we have
$$
4a^2x^2+4abx+4ac=\left(2ax+b+\sqrt{b^2-4ac}\ \right)\left(2ax+b-\sqrt{b^2-4ac}\ \right).\tag5
$$
Final step, dividing $(5)$ by $4a$ yields
$$
ax^2+bx+c=\color{blue}{\frac{\left(2ax+b+\sqrt{b^2-4ac}\ \right)\left(2ax+b-\sqrt{b^2-4ac}\ \right)}{4a}}.
$$
The process might look complicated, but once you understand the logic behind the process, especially for $(4)$, it will not be necessary anymore to memorize every step and your hand will automatically drive you to factorize every quadratic equation. 
