Factoring Quadratic Equations
Factoring is a method for solving quadratic equations.
This post is part of a series.
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 (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 $0$.
How to Factor
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 $(x+1)(x-3)$.
Case of Many Potential Factors
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.
Case of Negative Terms
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 $(x+4)(x-36)$.
Case of a Common Factor
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!
Case when the Leading Coefficient is Not One
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 chapter, when we cover the quadratic formula.
Case of No Middle Term
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.
Exercises
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:
$(7x-2)(3x+5)=0$
$x=\frac{2}{7} \text{ or } x=-\frac{5}{3}$
This post is part of a series.