# 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.

*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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