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 $0$.

$\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 2+x^2=-3x \\ \text{Convert to standard form} \hspace{.5cm} &\Bigg| \hspace{.5cm} x^2+3x+2=0 \\ \text{Factor} \hspace{.5cm} &\Bigg| \hspace{.5cm} (x+1)(x+2)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} x+1=0 \text{ or } x+2=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=-1 \text{ or } x=-2 \end{align*}$


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:

$\begin{align*} (x+1)(x+2) &= 0 \\ x(x+2) + 1(x+2) &= 0 \\ x^2 + 2x + 1x + 2 &= 0 \\ x^2 + 3x + 2 &= 0 \end{align*}$


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 $(x+8)(x+18)$.

$\begin{align*} \textbf{Factor Pair} & \hspace{.5cm} \textbf{Sum} \\ \text{1 and 144} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{145} \\ \text{2 and 72} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{74} \\ \text{3 and 48} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{51} \\ \text{4 and 36} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{40} \\ \text{6 and 24} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{30} \\ \textbf{8}\text{ and }\textbf{18} \hspace{.5cm} &\Bigg| \hspace{.5cm} \textbf{26} \\ \text{9 and 16} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{25} \\ \text{12 and 12} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{24} \end{align*}$


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 $(x-4)(x+36)$.

$\begin{align*} \textbf{Factor Pair} & \hspace{.5cm} \textbf{Difference} \\ \text{1 and 144} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{143} \\ \text{2 and 72} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{70} \\ \text{3 and 48} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{45} \\ \textbf{4}\text{ and }\textbf{36} \hspace{.5cm} &\Bigg| \hspace{.5cm} \textbf{32} \\ \text{6 and 24} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{18} \\ \text{8 and 18} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{10} \\ \text{9 and 16} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{7} \\ \text{12 and 12} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{0} \end{align*}$


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.

$\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3x^2-15+18=0 \\ \text{Factor a } 3 \text{ out of all terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3(x^2-5x+6)=0 \\ \text{Factor the quadratic expression} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3(x-2)(x-3)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} x-2=0 \text{ or } x-3=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=2 \text{ or } x=3 \end{align*}$


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

$\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} x^2+7x=0 \\ \text{Factor an } x \text{ out of all terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} x(x+7)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} x=0 \text{ or } x+7=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=0 \text{ or } x=-7 \\ \end{align*}$


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

$\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 2x^2-5x=0 \\ \text{Factor an } x \text{ out of all terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} x(2x-5)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} x=0 \text{ or } 2x-5=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=0 \text{ or } x=\frac{5}{2} \end{align*}$


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.

$\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 6x^2+11x+3=0 \\ \begin{matrix} \text{Multiply } 3 \text{ by } 6 \text{, and} \\ \text{replace } 6 \text{ with } 1 \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} x^2+11x+18=0 \\ \text{Factor normally} \hspace{.5cm} &\Bigg| \hspace{.5cm} (x+9)(x+2)=0 \\ \begin{matrix} \text{Divide each constant in} \\ \text{each factor by } 6 \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} \left( x+\frac{9}{6} \right) \left( x+\frac{2}{6} \right)=0 \\ \text{Simplify} \hspace{.5cm} &\Bigg| \hspace{.5cm} \left( x+\frac{3}{2} \right) \left( x+\frac{1}{3} \right)=0 \\ \begin{matrix} \text{Move denominators onto} \\ \text{variables} \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} (2x+3)(3x+1)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} 2x+3=0 \text{ or } 3x+1=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=-\frac{3}{2} \text{ or } x=-\frac{1}{3} \end{align*}$


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.

$\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3x^2-48=0 \\ \text{Factor } 3 \text{ out of all terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3(x^2-16)=0 \\ \text{Factor the quadratic} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3(x+4)(x-4)=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=-4 \text{ or } x=4 \end{align*}$



Practice Problems


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

$1) \hspace{.5cm} x^2+7x+12=0$
Solution:
$(x+3)(x+4)=0$
$x=-3 \text{ or } x=-4$

$2) \hspace{.5cm} x^2+9x+14=0$
Solution:
$(x+7)(x+2)=0$
$x=-7 \text{ or } x=-2$

$3) \hspace{.5cm} x^2-7x=-10$
Solution:
$(x-2)(x-5)=0$
$x=2 \text{ or } x=5$

$4) \hspace{.5cm} x^2+18=9x$
Solution:
$(x-6)(x-3)=0$
$x=6 \text{ or } x=3$

$5) \hspace{.5cm} x^2+2x=8$
Solution:
$(x+4)(x-2)=0$
$x=-4 \text{ or } x=2$

$6) \hspace{.5cm} 21-4x=x^2$
Solution:
$(x+7)(x-3)=0$
$x=-7 \text{ or } x=3$

$7) \hspace{.5cm} 3x+10=x^2$
Solution:
$(x+2)(x-5)=0$
$x=-2 \text{ or } x=5$

$8) \hspace{.5cm} x^2-5x=36$
Solution:
$(x+4)(x-9)=0$
$x=-3 \text{ or } x=9$

$9) \hspace{.5cm} 4x^2=-52x$
Solution:
$4x(x+13)=0$
$x=0 \text{ or } x=-13$

$10) \hspace{.5cm} -8x^2+64x=0$
Solution:
$8x(x-8)=0$
$x=0 \text{ or } x=8$

$11) \hspace{.5cm} x^2-25=0$
Solution:
$(x+5)(x-5)=0$
$x=-5 \text{ or } x=5$

$12) \hspace{.5cm} x^2-144$
Solution:
$(x+12)(x-12)=0$
$x=-12 \text{ or } x=12$

$13) \hspace{.5cm} 12x^2+11x=-2$
Solution:
$(4x+1)(3x+2)=0$
$x=-\frac{1}{4} \text{ or } x=-\frac{2}{3}$

$14) \hspace{.5cm} 10x^2=27x-5$
Solution:
$(2x-5)(5x-1)=0$
$x=\frac{5}{2} \text{ or } x=\frac{1}{5}$

$15) \hspace{.5cm} 5x=4-6x^2$
Solution:
$(3x+4)(2x-1)=0$
$x=-\frac{4}{3} \text{ or } x=\frac{1}{2}$

$16) \hspace{.5cm} 21x^2-10=-29x$
Solution:
$(7x-2)(3x+5)=0$
$x=\frac{2}{7} \text{ or } x=-\frac{5}{3}$

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