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 $4x^2 + 16x - 19$ ?

First of all, to find the roots of $4x^2+16x-19$ we have to calculate the discriminant:
If the second degree polynomial is of the form $$ax^2+bx+c=0$$ the discriminant is given from the formula: $$\Delta=b^2-4 \cdot a \cdot c$$
So the discriminant in this case is the following: 
$4x^2+16x-19=0 \Rightarrow \Delta=16^2-4 \cdot 4 \cdot (-19)=256+304 \Rightarrow \Delta=560$
Then the solutions are given from the formula: 
$$x_{1.2}=\frac{-b \pm \sqrt{\Delta}}{2 \cdot a}$$
Therefore we have the following solutions:
$x_{1,2}=\frac{-16 \pm \sqrt{560}}{2\cdot 4}=\frac{-16 \pm \sqrt{560}}{8}$
$x_1=\frac{-16-4 \sqrt{35}}{8}=-2-\frac{\sqrt{35}}{2} \text{ and } x_2=\frac{-16 +4\sqrt{35}}{8}=-2+\frac{\sqrt{35}}{2}$
Knowing that $$ax^2+bx+c=0 \Rightarrow a(x-x_1)(x-x_2)=0$$
we have the following:
$$4x^2+16x-19=4 \left ( x-  \left (-2-\frac{\sqrt{35}}{2}  \right ) \right ) \left ( x- \left (-2+\frac{\sqrt{35}}{2} \right ) \right )$$
