Factoring Quadratic Equations

Factoring is a method for solving quadratic equations. It involves converting the quadratic equation to standard form, then factoring it into a product of two linear terms (which are called factors), and finally solving for the variable values that make either factor equal to $0$.

When we factor, we are rearranging the equation to say that the product of two numbers is $0$. The equation is solved when either number is $0$, because any number multiplied by $0$ is $0$.

Factoring is easiest in hindsight. Multiplying through, we see that the factored form is equivalent to the standard form:

But how can we know this to begin with? In other words, if we want to factor an expression $x^2+bx+c$ into the form $(x+m)(x+n)$, how do we know what $m$ and $n$ are?

Here’s the trick: $m$ and $n$ need to multiply to $c$ and add to $b$.

To factor the expression $x^2+5x+4$, we need to find two numbers that multiply to $4$ and add to $5$. Although $2$ and $2$ multiply to $5$, they don’t add to $-2$. But $1$ and $-3$ multiply to $-3$ AND add to $-2$, so they work! The factored form is then $(x+1)(x-3)$.

Even with negatives, the method is still the same: to factor the expression $x^2-2x-3$, we need to find two numbers that multiply to $-3$ and add to $-2$. Although $-1$ and $3$ multiply to $-3$, they don’t add to $-2$. But $1$ and $-3$ multiply to $-3$ AND add to $-2$, so they work! The factored form is then $(x+1)(x-3)$.

Factoring can become a little tricky when $c$ has a lot of factors. In such cases, it can be helpful to make a factor table. For example, to factor $x^2+26x+144$, we can list out the factors of $144$ and find which pair adds to $26$. Since this pair is $8$ and $18$, the expression factors to $(x+8)(x+18)$.

To speed up the process, notice that the sums are automatically ordered from biggest to smallest – so we don’t necessarily have to create the whole table. We could have started with some intermediate pair, say $6$ and $24$, and realized that since the sum is too big, we need the first factor to be bigger than $6$. Or, we could have noticed that sum of $12$ and $12$ is in the ballpark of $26$, and worked our way up from the bottom of the table.

To deal with a negative value for $b$, we could use the same method as before, except that we would have to make both factors negative. For example, since we know that $8$ and $18$ are factors of $144$ that add to $26$, we also know that $-8$ and $-18$ are factors of $144$ that add to $-26$, so the expression $x^2-26x+144$ factors to $(x-8)(x-18)$.

To deal with a negative value for $c$, we can think about the difference instead of the sum. For example, to factor $x^2+32x-144$, we can find which factor pair of $144$ has a difference of $32$, and put a negative on the smaller factor to make the sum. Since this pair is $4$ and $36$, the expression factors to $(x-4)(x+36)$.

If $b$ were negative as well – say, if we wanted to factor $x^2-32x-144$ – then we could use the same process but put the negative on the bigger factor to make the sum negative. That is, we would put the negative on the $36$ instead of the $4$, and the resulting factored form would then be $(x+4)(x-36)$.

Sometimes, we can simplify quadratic expressions by factoring out something that ALL the terms have in common.

This makes it easy to factor quadratic expressions where $c$ is $0$ – just factor out the variable!

Factoring out the variable works even when $a$ is something other than $1$.

But what about when $a$ is something other than $1$, and $c$ is not zero?

There’s a little trick that lets us reduce this to a factoring problem with $a$ equal to $1$. We multiply $c$ by $a$, replace $a$ with $1$, factor the result, divide each constant in each factor by the original $a$, and move denominators onto our variables.

We’ll talk about why this trick works in the next section, when we cover the quadratic formula.

Lastly, what about when $b$ is $0$? Since the factors have to add to $b$, they must be negatives of each other. Since the factors have to multiply to $c$, and they are the same number (except one is negative), they must be the positive and negative square roots of $c$!

For example, $x^2-4$ factors to $(x+2)(x-2)$, and $x^2-9$ factors to $(x+3)(x-3)$.

This trick also works if $a$ is not equal to $1$ – we just have to factor $a$ out first.

Factor the following quadratic equations. Then, use the factored form to find the solutions. (You can view the solution by clicking on the problem.)