# Factoring

Sometimes it is possible to factor the quadratic equation ax^2+bx+c=0 as (x+p)(x+q)=0 . This requires a bit of practice at first but it is a very quick and effective way to find the solutions of a quadratic equation. The method uses the fact that if AB=0 then either A=0 or B=0 .

*So in the case of (x+p)(x+q) =0 , either (x+p)=0 or (x+q)=0 which means that x=-p or x=-q .*

The trick to factoring is to use the sum-product method. To illustrate this let’s find the solutions of 2x^2+6x-8=0 . First divide all terms by 2 to get x^2+3x-4=0 . To solve this equation we need to find two numbers which sum up to 3 and result in -4 when multiplied. In other words, p+q=3 and p\cdot q=-4 . By inspection we find p=4 and q=-1 .

The full solution to the quadratic equation 2x^2+6x-8=0 is:

\begin{aligned} 2x^2+6x-8=0 \\ x^2+3x-4=0 \\ (x+4)(x-1)=0 \\ x+4=0 ~\text{or}~ x-1=0 \\ x=-4 ~\text{or}~ x=1 \end{aligned}