Factoring Tip!

I figured this out when I wasn't paying attention in math

if you want to split $\displaystyle z_1x^2+z_2x+z_3$ into $\displaystyle (ax+b)(cx+d)$

You can split it into $\displaystyle z_1x^2+\left(\frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}\right)x+\left(\frac{z_2-\sqrt{z_2^2-4z_1z_3}}{2}\right)x+z_3$

note that either $\displaystyle \frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}=ad$ or $\displaystyle \frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}=bc$

let's say we have: $\displaystyle -6a^2 - 5a -1$

So you can split it into this: $\displaystyle z_1x^2+\left(\frac{z_2+\sqrt{z_2^2-4z_1z_3}}{2}\right)x+\left(\frac{z_2-\sqrt{z_2^2-4z_1z_3}}{2}\right)x+z_3$

substitute: $\displaystyle -6x^2+\left(\frac{-5+\sqrt{(-5)^2-4(-6)(-1)}}{2}\right)x+$$\displaystyle \left(\frac{-5-\sqrt{(-5)^2-4(-6)(-1)}}{2}\right)x-1$

thus: $\displaystyle -6x^2+\left(\frac{-5+\sqrt{25-24}}{2}\right)x+\left(\frac{-5-\sqrt{25-24}}{2}\right)x-1$

therefore: $\displaystyle -6x^2+\left(\frac{-5+\sqrt{1}}{2}\right)x+\left(\frac{-5-\sqrt{1}}{2}\right)x-1$

thus: $\displaystyle -6x^2+\left(\frac{-5+1}{2}\right)x+\left(\frac{-5-1}{2}\right)x-1$

then: $\displaystyle -6x^2+\left(\frac{-4}{2}\right)x+\left(\frac{-6}{2}\right)x-1$

then: $\displaystyle -6x^2-2x-3x-1$

note that it is now in either the form: $\displaystyle acx^2+adx+bcx+bd$ or $\displaystyle acx^2+bcx+adx+bd$