Factoring

# 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}