How To Factor Quadratic Equations

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How to factor quadratic equations

Here you will learn how to factor quadratic equations.

Students will first learn how to factor quadratic equations as part of geometry in high school.

What is it to factor a quadratic equation?

To factor a quadratic equation means to determine the two linear factors of the quadratic function. It simplifies the quadratic equation and makes it easier to find the roots.

Quadratic equations are a type of polynomial equation because they consist of algebraic terms, with the highest term being second-degree $($ the exponent of the highest $x$ term is $2).$

The standard form of the quadratic equation is:

$a{x}^2+bx+c=0$

Let’s look at how to factor a quadratic equation.

For example,

x^2+9 x+16=- \, 2

When you solve a quadratic equation, make sure the equation is set equal to zero first. This is important because of the zero product property, which helps you find the solutions; if $ab=0$ then $a=0$ or $b=0.$

Add $2$ to both sides of $x^2+9 x+16=- \, 2$ so that the equation equals $0.$

The quadratic will factor into the product of two linear expressions, in the form of the two pairs of parentheses:

List the factor pairs of the $c$ term (the constant term):

1, \, 18

2, \, 9

3, \, 6

The $b$ term (the term including the coefficient of $x$ ) is $9,$ so look for the factor pairs that multiply to $18$ and sum to $9.$

The factors of $3$ and $6$ multiply to $18$ and add to $9.$ Place each of these into the parentheses to get $(x+3)(x+6).$

$x^2+9 x+16=- \, 2$ or $x^2+9 x+18=0$ written in its factored form is $(x+3)(x+6)=0.$

Keep in mind that the factors can be multiplied in any order. So, $(x+6)(x+3)$ is the same as $(x+3)(x+6)$ since both expand to be $x^2+9 x+18.$

What is it to factor a quadratic equation?

Common Core State Standards

How does this relate to high school math?

Algebra – Seeing Structure in Expressions (HSA.SSE.B.3a)
Factor a quadratic expression to reveal the zeros of the function it defines.

[FREE] Factoring Quadratic Equations Worksheet

Use this worksheet to check your students’ understanding of factoring quadratic equations. 15 questions with answers to identify areas of strength and support!

DOWNLOAD FREE

[FREE] Factoring Quadratic Equations Worksheet

Use this worksheet to check your students’ understanding of factoring quadratic equations. 15 questions with answers to identify areas of strength and support!

DOWNLOAD FREE

In order to factor a quadratic equation when the coefficient of the $a$ term is $1\text{:}$

Find two factors of the constant, $\textbf{c}$ term, that sum to equal the coefficient of the $\textbf{b}$ term.
Write the quadratic in factored form with two sets of parentheses.
Check your work and write the quadratic equation in factored form.

How to factor quadratic equation examples

Example 1: factoring a trinomial

Factor the quadratic expression.

x^2-2 x-15

Find two factors of the constant, $\textbf{c}$ term, that sum to equal the coefficient of the $\textbf{b}$ term.

The constant, $c$ term, is $-15,$ so one factor must be negative. The factors of $-15$ are:

\begin{aligned} 1, & ~-15 \\\\ -1, & ~15 \\\\ 3, & ~-5 \\\\ -5, & ~3 \end{aligned}

The coefficient of the $b$ term is $- \, 2.$

The two factors that multiply to $- \, 15$ and add to $- \, 2$ are $3$ and $- \, 5.$

\begin{aligned}& 3 \times(- \, 5)=- \, 15 \\\\ & 3+(- \, 5)=- \, 2 \end{aligned}

2Write the quadratic in factored form with two sets of parentheses.

The $a$ term breaks up to be $x\cdot{x}$ because $x\cdot{x}=x^{2}.$

So, $x^2-2 x-15=(x+3)(x-5).$

3Check your work and write the quadratic equation in factored form.

You can check your work by calculating the partial products with an area model.

x^2-5 x+3 x-15=x^2-2 x-15

Example 2: factoring a trinomial

Factor the quadratic expression.

x^2+14 x+45

Find two factors of the constant, $\textbf{c}$ term, that sum to equal the coefficient of the $\textbf{b}$ term.

The constant, $c$ term, is $45$ and the $b$ term is positive, so both factors must be positive. The factors of $45$ are:

$\begin{aligned}& 1,~45 \\\\ & 5,~9 \\\\ & 3,~15 \end{aligned}$

The coefficient of the $b$ term is $14.$

The two factors that multiply to $45$ and add to $14$ are $5$ and $9.$

$\begin{aligned}& 5 \times 9=45 \\\\ & 5+9=14 \end{aligned}$

Write the quadratic in factored form with two sets of parentheses.

The $a$ term breaks up to be $x\cdot{x}$ because $x\cdot{x}=x^{2}.$

So, $x^2+14 x+45=(x+9)(x+5).$

Check your work and write the quadratic equation in factored form.

You can check your work by calculating the partial products with an area model.

$x^2+9 x+5 x+45=x^2+14 x+45$

Example 3: factoring a binomial

Factor the quadratic expression.

x^2-49

Find two factors of the constant, $\textbf{c}$ term, that sum to equal the coefficient of the $\textbf{b}$ term.

Factoring binomials has the same steps as factoring trinomials. The constant, $c$ term, is $- \, 49$ and the coefficient of the $b$ term is $0.$ This means the two factors are opposites. The factors of $- \, 49$ that are opposites are:

$\begin{aligned} 1, & ~-49 \\\\ -1, & ~49 \\\\ 7, & ~-7~\checkmark \end{aligned}$

Notice how these factors connect to the square roots of $49.$

These factors multiply to $- \, 49$ and add to $0.$

$\begin{aligned}& 7 \times(- \, 7)=- \, 49 \\\\ & 7+(- \, 7)=0 \end{aligned}$

Write the quadratic in factored form with two sets of parentheses.

The $a$ term breaks up to be $x\cdot{x}$ because $x\cdot{x}=x^{2}.$

So, $x^2-49=(x-7)(x+7).$

Check your work and write the quadratic equation in factored form.

You can check your work by calculating the partial products with an area model.

$x^2+(-7 x)+7 x-49=x^2-49$

Example 4: factoring a perfect square

Factor the quadratic expression.

x^2+22 x+121

Find two factors of the constant, $\textbf{c}$ term, that sum to equal the coefficient of the $\textbf{b}$ term.

The constant, $c$ term, is $121$ and the $b$ term is positive, so both factors must be positive. The factors of $121$ are:

$\begin{aligned}& 1,~121 \\\\ & 11,~11 \end{aligned}$

The coefficient of the $b$ term is $22.$

The two factors that multiply to $121$ and add to $22$ are $11$ and $11.$

$\begin{aligned}&11\times11=121 \\\\ &11+11=22 \end{aligned}$

Write the quadratic in factored form with two sets of parentheses.

The $a$ term breaks up to be $x\cdot{x}$ because $x\cdot{x}=x^{2}.$

So, $x^2+22 x+121=(x+11)(x+11).$

Check your work and write the quadratic equation in factored form.

You can check your work by calculating the partial products with an area model.

$x^2+11 x+11 x+121=x^2+22 x+121$

In order to factor a quadratic equation when the $a$ term is not $1\text{:}$

Find the factors of $\textbf{ac}$ that sum to equal the coefficient of the $\textbf{b}$ term.
Split up the $\textbf{x}$ term and using grouping to factor.
Check your work and write the quadratic equation in factored form.

Example 5: factoring quadratic equations with a leading coefficient ≠ 1

Factor the quadratic, $3x^2+28x+32.$

Find the factors of $\textbf{ac}$ that sum to equal the coefficient of the $\textbf{b}$ term.

The value of $ac$ is $3 \times 32=96.$

The factors of $96$ are…

$\begin{aligned}&1,~96 \\\\ &2,~48 \\\\ &3,~32 \\\\ &4,~24 \\\\ &6,~16 \\\\ &8,~12 \end{aligned}$

The coefficient of the $b$ term is $28.$ The factors of $96$ that sum to $28$ are $4$ and $24.$

Split up the $\textbf{x}$ term and using grouping to factor.

Replace $28x$ with $4x +24x$ in the quadratic equation: $3 x^2+4 x+24 x+32$

Now, group the first two and last two terms: $\left(3 x^2+4 x\right)+(24 x+32)$

Then, factor out the greatest common factor in each: $x(3 x+4)+8(3 x+4)$

Step-by-step guide: Factor by grouping

Check your work and write the quadratic equation in factored form.

x(3 x+4)+8(3 x+4)=(x+8)(3 x+4)

You can check your work by calculating the partial products with an area model.

$3 x^2+24 x+4 x+32=3 x^2+28 x+32$

Example 6: factoring quadratic equations with a leading coefficient ≠ 1

Factor the quadratic, $7 x^2+40 x-12.$

Find the factors of $\textbf{ac}$ that sum to equal the coefficient of the $\textbf{b}$ term.

The value of $ac$ is $7 \times(- \, 12)=- \, 84.$

The factors of $- \, 84$ are…

$\begin{aligned}1, & ~- 84 \\\\ -1,& ~84 \\\\ 2,& ~-42 \\\\ -2,& ~42 \\\\ 3,& ~-28 \\\\ -3,& ~28 \\\\ 4,& ~-21 \\\\ -4,& ~21 \\\\ 6,& ~-14 \\\\ -6,& ~14 \\\\ 7,& ~-12 \\\\ -7,& ~12 \end{aligned}$

The coefficient of the $b$ term is $40.$ The factors of $- \, 84$ that sum to $40$ are $- \, 2$ and $42.$

Split up the $\textbf{x}$ term and using grouping to factor.

Replace $40x$ with $42x+(- \, 2x)$ in the quadratic equation:

$7 x^2+42 x+(- \, 2 x)-12$

Now, group the first two and last two terms: $\left(7 x^2+42 x\right)+(- \, 2 x-12)$

Then, factor out the greatest common factor in each: $7 x(x+6)-2(x+6)$

Check your work and write the quadratic equation in factored form.

7 x(x+6)-2(x+6)=(7 x-2)(x+6)

You can check your work by calculating the partial products with an area model.

$7 x^2+42 x+(- \, 2 x)-12=7 x^2+40 x-12$

Teaching tips for how to factor quadratic equations

Review factoring, GCF and how to find common factors before teaching students how to factor quadratic equations.

When factoring quadratic, start with simple quadratics that do not require the quadratic formula, as models on this page.

It is also important to expose students to different methods of factoring and give them opportunities to decide which methods are more efficient for certain equation types.

Students should also be encouraged to check their work with a different method than they solved with.

Look for activities and worksheets that have students factor quadratic equations, but not solve them. Students do not need to know how to solve quadratic equations in order to factor them.

Easy mistakes to make

Thinking that quadratic equations only look like $\textbf{ax}\bf{^2} \textbf{ +} \textbf{ bx + c} \textbf{ = } \bf{ 0 }$
A quadratic equation is any equation in which the highest degree term is $x^2$ (a second-degree polynomial). While quadratic equations don’t always start in the form $a x^2+b x+c=0,$ they can always be simplified into this standard form.

For example,
Simplify $5 x^2-3 x=2 x^2+14$ to the form $ax^2+bx+c=0.$

Confusing the middle term and constant term when factoring
As the step-by-step guide indicates, it is important to use the constant term $c \, ($ and sometimes $a)$ to identify possible factors.

Then the middle term $b$ is used to identify which two factors are used to factor the quadratic. Confusing these terms or this process leads to incorrect answers.

Thinking quadratics must always be simplified on the left side
While the left side of the equation is most commonly used, you will get the exact same results by factoring on the right side.

For example, $0=x^2-2 x-15$ factors to $0=(x+3)(x-5),$ just as it does on the left side of the equation or in an expression, as shown in Example $1.$

Practice how to factor quadratic equations questions

(x-4)(x+7)

(x+4)(x-7)

(x+2)(x-14)

(x+14)(x-2)

The constant, $c$ term, is $– \, 28,$ so one factor must be negative. The factors of $– \, 28$ are:

\begin{aligned} 1,& ~-28 \\\\ -1,& ~28 \\\\ 2,& ~14 \\\\ -2,& ~14 \\\\ 4,& –7 \\\\ -4,& ~7 \end{aligned}

The coefficient of the $b$ term is $– \, 3.$

The two factors that multiply to $– \, 28$ and add to $– \, 3$ are $4$ and $– \, 7.$

\begin{aligned}4 \times(-7)& =-28 \\\\ 4+(-7)&=-3 \end{aligned}

The $a$ term breaks up to be $x\cdot{x}$ because $x\cdot{x}=x^{2}.$

So, $x^2-3 x-28=(x+4)(x-7).$

You can check your work by calculating the partial products with an area model.

How to factor quadratic equations 11 US

x^2+4 x-7 x-28=x^2-3 x-28

(x+3)(x+6)

(x-3)(x-6)

(x-2)(x-9)

(x+2)(x+9)

The constant, $c$ term, is $18$ and the $b$ term is positive, so both factors must be positive. The factors of $18$ are:

\begin{aligned}& 1,~18 \\\\ & 2,~9 \\\\ & 3,~6 \end{aligned}

The coefficient of the $b$ term is $11.$

The two factors that multiply to $18$ and add to $11$ are $2$ and $9.$

\begin{aligned}2 \times 9&=18 \\\\ 2+9&=11 \end{aligned}

The $a$ term breaks up to be $x\cdot{x}$ because $x\cdot{x}=x^{2}.$

So, $x^2+11 x+18=(x+2)(x+9).$

You can check your work by calculating the partial products with an area model.

How to factor quadratic equations 12 US

x^2+2 x+9 x+18=x^2+11 x+18

(x-20)(x+20)

(x-20)(x-20)

(x-40)(x+10)

(x-40)(x-10)

The constant, $c$ term, is $– \, 400$ and the coefficient of the $b$ term is $0.$ This means the two factors are opposites.

The factors of $– \, 400$ that are opposites are:

– \, 20, \, 20

Note: You could have listed all $16$ factor pairs and then solved, but realizing that you only want factor pairs that sum to $0$ before you list all the factor pairs, saves time!

These factors multiply to $– \, 400$ and add to $0.$

\begin{aligned} 20 \times(-20)& =-400 \\\\ 20+(-20)& =0 \end{aligned}

The $a$ term breaks up to be $x\cdot{x}$ because $x\cdot{x}=x^{2}.$

So, $x^2-400=(x-20)(x+20).$

You can check your work by calculating the partial products with an area model.

How to factor quadratic equations 13 US

x^2+20 x+(- \, 20 x)-400=x^2-400

(x+1)(x+15)

(x-1)(x-15)

(x-3)(x-5)

(x+3)(x+5)

The constant, $c$ term, is $15$ and the coefficient of the $b$ term is negative. This means the two factors are negatives.

The negative factors of $15$ are:

\begin{aligned}-1,& ~-15 \\\\ -3,& ~-5 \end{aligned}

The coefficient of the $b$ term is $– \, 8.$

The two factors that multiply to $15$ and add to $– \, 8$ are $– \, 3$ and $– \, 5.$

\begin{aligned} -3 \times(-5)&=15 \\\\ -3+(-5)&=-8 \end{aligned}

The $a$ term breaks up to be $x\cdot{x}$ because $x\cdot{x}=x^{2}.$

So, $x^2-8 x+15=(x-3)(x-5).$

You can check your work by calculating the partial products with an area model.

How to factor quadratic equations 14 US

x^2+(- \, 3 x)+(- \, 5 x)+15=x^2-8 x+15

(x+3)(9x+10)

(x+30)(9x+9)

(9x+10)(x+27)

(9x+5)(x+6)

The value of $ac$ is $9 \times 30=270.$

The factors of $270$ are…

\begin{aligned}& 1,~270 \\\\ & 2,~135 \\\\ & 3,~90 \\\\ & 5,~54 \\\\ & 6,~45 \\\\ & 9,~30 \\\\ & 10,~27 \\\\ & 15, 18 \end{aligned}

The coefficient of the $b$ term is $37.$ The factors of $270$ that sum to $37$ are $10$ and $27.$

Replace $37x$ with $10x+27x$ in the quadratic equation:

9 x^2+10 x+27 x+30

Now, group the first two and last two terms: $\left(9 x^2+10 x\right)+(27 x+30)$

Then, factor out the greatest common factor in each:

x(9 x+10)+3(9 x+10)=(x+3)(9 x+10)

You can check your work by calculating the partial products with an area model.

How to factor quadratic equations 15 US

9 x^2+27 x+10 x+30=9 x^2+37 x+30

(2x-3)(2x-3)

(2x+3)(2x-3)

(x-3)(2x+3)

(x+3)(2x-3)

The value of $ac$ is $2 \times(-9)=-18.$

The factors of $-18$ are…

\begin{aligned} 1,& ~-18 \\\\ -1,& ~18 \\\\ 2,& ~-9 \\\\ -2,& ~9 \\\\ 3,& ~-6 \\\\ -3,& ~6 \end{aligned}

The coefficient of the $b$ term is $– \, 3.$ The factors of $– \, 18$ that sum to $– \, 3$ are $– \, 6$ and $3.$

Replace $– \, 3x$ with $3x +(- \, 6x)$ in the quadratic equation:

2 x^2+3 x+(- \, 6 x)-9

Now, group the first two and last two terms: $\left(2 x^2+3 x\right)+(- \, 6 x-9)$

Then, factor out the greatest common factor in each:

x(2 x+3)-3(2 x+3)=(x-3)(2 x+3)

You can check your work by calculating the partial products with an area model.

How to factor quadratic equations 16 US

2 x^2-6 x+3 x-9=2 x^2-3 x-9

How to factor quadratic equation FAQs

What is factorising?

It is another version of the term as ‘factoring’ and it has the same meaning.

What are possible values for $\textbf{a, b}$ or $\textbf{c}$ in a quadratic equation?

These variables can have a value of any real numbers, including fractions, negative numbers or irrational numbers.

Can you factor any quadratic polynomial?

Yes, either with rational numbers or with irrational or complex numbers.

What math classes teach this topic?

This topic can appear in precalculus, Algebra $I,$ Algebra $II$ or other integrated math classes.

Where does the term “quadratic” come from?

The term is based on the Latin word quadratum, which means square. The highest term in a quadratic equation is $x$ multiplied by itself or squared. Therefore the word “quadratic” was adapted to use in English to refer to this type of equation.

The next lessons are

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