<s>[INST] 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 $4x^2 + 16x - 19$ ?
 [/INST] If we multiply two monomials
$$(x-a)(x-b)=x^2-(a+b)x+ab$$
Think of this in terms of the arithmetic mean of the roots $\mu=(a+b)/2$ and the geometric mean $\gamma=\sqrt{ab}$:
$$x^2-2\mu+\gamma^2$$
Then
$$\begin{align}\mu^2-\gamma^2& =\left(\frac{a+b}{2}\right)^2-ab\\
& =\frac{a^2+2ab+b^2}{4}-\frac{4ab}{4}\\
& =\frac{a^2-2ab+b^2}{4}\\
& =\left(\frac{a-b}{2}\right)^2\\
& =\rho^2
\end{align}$$
Where $\rho=(a-b)/2$ is the "radius" from the mean $\mu$ to the roots $a,b$. We can recover the roots from
$$\{a,b\}=\mu\pm\rho=\mu\pm\sqrt{\mu^2-\gamma^2}$$
Taking your example, we see that $4x^2+16x−19=4(x^2+4x-19/4)=4(x^2-2(-2)x-19/4)$ so $\mu=-2$ and $\gamma=-19/4$, giving us the roots
$$\begin{align}
\{a,b\}& =-2\pm\sqrt{(-2)^2-(-19/4)}\\
& =-2\pm\sqrt{4+19/4}\\
& =-2\pm\sqrt{35/4}\\
& =-2\pm\sqrt{35}/2
\end{align}$$
If you need to, you can then rewrite the original expression as
$$\begin{align}
4x^2+16x−19& =4(x-(-2+\sqrt{35}/2))(x-(-2-\sqrt{35}/2))\\
\quad& =4(x+2-\sqrt{35}/2)(x+2+\sqrt{35}/2)\end{align} 
$$
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