To factor an expression means to rewrite it so that it is the product of factors. For example, the expression $x^2-7x+12$ can be written as $(x-3)(x-4)$. 

When we multiply $(x-3)$ times $(x-4)$ to obtain $x^2-7x+12$ we call that operation "multiplying out" or sometimes FOILing. (Recall that FOIL stands for First, Outer, Inner, Last, which is how we combine the terms.) The reverse process is called factoring. 

Solving Quadratic Equations by Factoring

Factoring is useful to help solve an equation of the form:

$ax^2+bx+c=0$ 

For example, if you wanted to solve the equation $x^2-7x+12=0$, if you could realize that the quadratic factors as $(x-3)(x-4)$.

You could then rewrite your equation as $(x-3)(x-4)=0 $ and conclude that the two solutions are $x=3$ and $x=4$. 

This follows because the only way that two things can multiply together to be $0$ is if one or the other is $0$. So the expression $(x-3)(x-4)=0$ tells us that either $x-3=0$, in which case $x=3$, or $x-4=0$, in which case $x=4.$

Determining the Factors

You may be wondering: but how does one come up with the two factors? Let's consider this same example more extensively.

Again, imagine you want to factor $x^2-7x+12$. First set up the desired format: $( \quad )( \quad )$

Now lets look at how to fill in the variable slots in this format. The first two terms in each set of parentheses must multiply together to be $x^2$. So we write $(x \quad )(x \quad )=0$. 

Next we think about the fact that the last two terms must multiply together to be $12.$ We consider the following possibilities for how two numbers could multiply together to be $12,$ namely $12 \cdot 1, 6 \cdot 2,$ or $ 4 \cdot 3.$ 

Of these possibilities, the ones that add to be $7$ (the middle coefficient) are $4$ and $3$. Since we are trying to get negative $7x$ in the middle, we try $(x-4)(x-3)$. FOILing gives exactly what we want, namely:

$\begin{aligned}(x-4)(x-3)&= x^2-3x-4x+12 \\ &=x^2-7x+12 \end{aligned}$ 

Note that we can (and should) check that $4$ and $3$ are solutions to the original equation $x^2-7x+12=0.$ 

Another Example

Suppose you want to solve:

$x^2+2x-8=0$ 

We attempt to factor the quadratic. We would need factors of $-8$ which also add together to be $2$. It seems as though $4$ and $-2$ might work, so we try factoring this as $(x-2)(x+4).$ 

FOILing shows us that this is correct:

$\begin{aligned} (x-2)(x+4) &= x^2 +4x-2x-8 \\ &= x^2 +2x-8 \end{aligned}$

Thus we have $(x-2)(x+4)=0,$ so either $x-2=0$ in which case $x=2,$ or $x+4=0,$ in which case $x=-4.$ We check our work by plugging both of these numbers back into the original equation and seeing that the equation is satisfied.