Partial Fractions

The roots of $ax^2 +bx+c$ can be represented as $\dfrac{-b+\sqrt{b^2 -4ac}}{2a}$ and $\dfrac{-b-\sqrt{b^2 -4ac}}{2a}$ . Thus, the partial fractions will be written like $\displaystyle\frac{A}{x- \dfrac{-b+\sqrt{b^2 -4ac}}{2a}} + \frac{B}{x- \dfrac{-b-\sqrt{b^2 -4ac}}{2a}}$ . Rewrite as $B \left( x- \dfrac{-b+\sqrt{b^2 -4ac}}{2a} \right) + A \left( x- \dfrac{-b-\sqrt{b^2 -4ac}}{2a}\right) =1$ . Substitute $\dfrac{-b+\sqrt{b^2 -4ac}}{2a}$ you will get $A\left(\dfrac{\sqrt{b^2 -4ac}}{a}\right)=1$ , which gives A as $\dfrac{a}{\sqrt{b^2 - 4ac}}$ . Do the same except substitute $\dfrac{-b-\sqrt{b^2 -4ac}}{2a}$ and you will get B as $-\dfrac{a}{\sqrt{b^2 - 4ac}}$ . The sum of A and B are zero, and you will see that I put a=2014, b=2015, and c=2016 for show.