Solving quadratic equations by factoring is one of the most efficient methods for finding the “roots” (solutions) of a quadratic equation. While not every equation can be solved using factoring, using this method whenever possible can save some time and also strengthen your factoring abilities.
In this article, we’ll explore different methods of factoring to solve simple and complex equations as well as look at some full examples with detailed answers. Let’s get started!
Return to the Table of Contents
What We Review
What is Factoring?
Before we dig deep into factoring quadratic equations, let’s remember what factors are by looking at numerical examples. We can use 12 as an example. The numbers 6 and 2 are factors of 12 because multiplying 6 and 2 gives the product of 12 .
There are other ways to factor 12, as well, such as using the factors 4 and 3 instead.
| Quick Reminders ● Factoring a number or expression means breaking it into separate factors. ● Factors are terms that, when multiplied together, produce the original number or expression. |
Likewise, when we factor the standard from of a quadratic equation:
y=ax^2+bx+c
…into factored form:
y=(x-r_1)(x-r_2)
…we are simply saying that when we multiply (x-r_1) and (x-r_2), we will get the product ax^2+bx+x.
Return to the Table of Contents
Factoring Quadratic Equations Examples
Before things get too complicated, let’s begin by solving a simple quadratic equation. When you are asked to “solve a quadratic equation”, you are determining the x-intercepts. The x-intercepts can also be referred to as zeros, roots, or solutions. All of these terms are the same. For our purpose, a simple quadratic means a quadratic where a=1. Remember, the standard form of a quadratic is:
y=ax^2+bx+c
For more information about forms of quadratics, check out our article on the different forms of quadratics.
If a=1, then no coefficient appears in front of x^2. In these cases, solving quadratic equations by factoring is a bit simpler because we know factored form, y=(x-r_1)(x-r_2), will also have no coefficients in front of x. We simply must determine the values of r_1 and r_2. But no need to worry, we include more complex examples in the next section.
Solving y=x^2+7x+12
To start, let’s try solving the equation:
y=x^2+7x+12
…where a=1, b=7, and c=12. To begin, we can set y=0. This gives us the equation x^2+7x+12=0. Now, let’s solve for the x-intercepts by factoring.
Find the Correct Factors of 12
To factor the expression x^2+7x+12, we must first determine the factors of c, in this case 12. There are many ways 12 can be factored, such as using -12 and -1 or using 6 and 2.
To determine which factors to use, we must also determine the sum of the factors. The sum should add up to b, in this case 7. Let’s make a table determining the sum of all the ways to factor 12.
| Factor Pairs of 12 | Sum |
|---|---|
| 12 and 1 | 13 |
| -12 and -1 | -13 |
| 6 and 2 | 8 |
| -6 and -2 | -8 |
| \color{blue}{4} and \color{blue}{3} | \color{blue}{7} |
| -4 and -3 | -7 |
The factor pair 4 and 3 multiplies to 12 and adds to 7. Therefore, we can factor x^2+7x+12=0 into:
(x+3)(x+4)=0
Solve for x
To solve for the value(s) of x, we must set each factor equal to 0. Remember, (x+3) is being multiplied with (x+4). Therefore, if (x+3)=0, then (x+3)(x+4)=0. This is because we know that anything multiplied by 0 is 0.
When we set both factors equal to zero, we obtain two equations: x+3=0 and x+4=0. We can now solve these equations.
| x+3=0 x=-3 | x+4=0 x=-4 |
Therefore, the two x-intercepts of the equation y=x^2+7x+12 are at -3 and -4.
We can verify by creating the graph of y=x^2+7x+12.
x-intercepts: x=-3 and x=-4
Return to the Table of Contents
Solving y=x^2-8x-20
Let’s determine the roots (or x-intercepts) of the equation:
y=x^2-8x-20
…where a=1, b=-8 and c=-20. To begin, let’s set y=0. This gives us the equation x^2-8x-20=0. Now, we solve by factoring.
Find the Correct Factors of -20
To factor the expression x^2-8x-20, we must first determine the factors of c (in this case -20). Again, there are many possible factor pairs. The sum should add up to b, in this case -8.
We know the factors must have opposite signs because a positive multiplied by a negative will produce a negative number. For example, (1)(-20)=-20. We also know the numbers must add up to a negative number. This means the number with the larger absolute value in the factor pair must be *negative*. We will only include such factor pairs in the table below.
| Relevant Factor Pairs of -20 | Sum of Factors |
|---|---|
| -20 and 1 | -19 |
| \color{blue}{-10} and \color{blue}{2} | \color{blue}{-8} |
| -5 and 4 | -1 |
The factor pair -10 and 2 multiplies to -20 and adds to -8. Therefore, we can factor x^2-8x-20=0 into (x-10)(x+2)=0.
Solve for x
To solve for the value(s) of x, we will set each factor equal to 0. When we set both factors equal to zero, we obtain two equations: x-10=0 and x+2=0. Let’s solve these equations.
| x-10=0 x=10 | x+2=0 x=-2 |
Therefore, the two x-intercepts of the equation y=x^2-8x-20 are at 10 and -2.
We can verify by creating the graph of y=x^2-8x-20.
x-intercepts: x=10 and x=-2
Remember: if a quadratic equation doesn’t appear factorable, you can always try solving it by using the quadratic formula.
Return to the Table of Contents
Solving Quadratic Equations with the “AC Method”
As promised, we can now demonstrate the more difficult cases where a \ne 1. Do not worry! Having a great understanding of the simpler method will make this approach very manageable. Just as builders must start construction with a strong foundation, our foundational knowledge of factoring prepares us for greater challenges. We are now ready to build our tower higher and factor the more difficult equations.
We can begin by solving the equation:
y=6x^2-7x-3
…where a=6, b=-7 and c=-3. First, let us set y equal to 0. This gives us the equation 6x^2-7x-3=0. Before we begin factoring, we will determine the value of ac. This is why this is called the ac method.
In this case, a=6 and c=-3-, so ac=(6)(-3)=-18.
Create Factor Pairs of -18
We will create our factor pairs using factors of -18. Just as before, the factor pairs should add up to the value of b (in this case, -7). Because the factors must multiply to a negative value and add to a negative value, we will only include factor pairs where the number with the larger absolute value is negative.
| Relevant Factor Pairs of -18 | Sum of Factors |
|---|---|
| -18 and 1 | -17 |
| \color{blue}{-9} and \color{blue}{2} | \color{blue}{-7} |
| -6 and 3 | -3 |
The factor pair -9 and 2 has a product of -18 and a sum of -7. We are not ready to factor just yet. We will rewrite the equation, 6x^2-7x-3=0, by changing -7x into -9x+2x. (Note: it would also be correct to write -7x as 2x-9x.)
6x^2-7x-3=0
6x^2-9x+2x-3=0
Now, we will use factor by grouping to factor this expression. To use factor by grouping, we will factor out the greatest common factor from the first two terms and factor out the greater common factor of the last two terms.
In the equation, 6x^2-9x+2x-3=0, the first two terms are 6x^2 and -9x. We can factor out 3x from both terms.
The second two terms are 2x and -3. These terms have no factors in common, other than 1. In this case, we factor out 1.
This changes the equation:
6x^2-9x+2x-3=0
…into:
3x(2x-3)+1(2x-3)=0
Now we see that (2x-3) is a common factor, so we can rewrite 3x(2x-3)+1(2x-3)=0 as:
(2x-3)(3x+1)=0
Solve for x
Yes! We have now factored our original equation 6x^2-7x-3=0 into (2x-3)(3x+1)=0. We must remember not to stop here, but to continue on until we have solved for the value(s) of x. We will set each factor equal to 0 and solve to obtain our solution(s).
| 2x-3=0 2x=3 x=\dfrac{3}{2} | 3x+1=0 3x=-1 x=-\dfrac{1}{3} |
Therefore, the x-intercepts of y=6x^2-7x-3 are at \dfrac{3}{2} and at -\dfrac{1}{3}.
We can verify by creating the graph of y=6x^2-7x-3.
x-intercepts: x=\dfrac{3}{2} and x=-\dfrac{1}{3}
Return to the Table of Contents
Video Examples of Factoring Quadratic Equation
In addition to the written examples above, here’s a detailed video with a few different examples of how to sole quadratic equations by factoring:
Factoring Trinomial with The Box Method
Often the challenges we face in mathematics are about how to organize the information we have. As problems become more complex, we need more methods of organizing our information. We know the right organizational system can make a huge difference when working through a project.
Organizing our information can also make a huge difference when solving more complex equations. The box method is very similar to the ac method we discussed above. If the ac felt like too much information to organize, the box method may help you keep track of this process.
Example Using The Box Method
We can demonstrate this method by solving the quadratic equation:
y=7x^2+26x-8
…where a=7, b=26, and c=-8. We begin again by setting y=0, giving us the equation 7x^2+26x-8=0.
We create a box cut into four parts. On the top left, we will put the first term, 7x^2, and in the bottom right, we will put the last term -8.
| 7x^2 | |
| -8 |
To determine the remaining two places in the box, we use the factor pair of ac with a sum of b. In this case, ac=(7)(-8)=-56 and b=26. We must determine the factor pair of -56 that has a sum of 26. Because the factor pair must multiply to a negative number and add to a positive number, we will only into the factor pairs where the number with the larger absolute value is positive.
| Relevant Factor Pairs of -56 | Sum of Factors |
|---|---|
| 56 and -1 | 55 |
| \color{blue}{28} and \color{blue}{-2} | \color{blue}{26} |
| 14 and -4 | 10 |
| 8 and -7 | 1 |
The factor pair 28 and -2 multiplies to -56 and adds to 26.
Now, we can fill in 28x and -2x into the remaining places in the box. The location of 28x and -2x can be interchanged. For our example, we will put 28x into the top right space and -2x into the bottom left space.
| 7x^2 | 28x |
| -2x | -8 |
After that, we will write the greatest common factors for each row on the left. The greatest common factor of 7x^2 and 28x is 7x. The greatest common factor of -2x and -8 is -2.
| 7x | 7x^2 | 28x |
| -2 | -2x | -8 |
Now, we will write the greatest common factors for each column on the top. The greatest common factor of 7x^2 and -2x is x. The greatest common factor of 28x and -8 is 4.
| x | 4 | |
| 7x | 7x^2 | 28x |
| -2 | -2x | -8 |
We use the greatest common factors we wrote as the factors of our equation. We can rewrite 7x^2+26x-8=0 as 0=(7x-2)(x+4).
Solve for x
Now, we can set each factor equal to 0 and solve for the solution(s).
| 7x-2=0 | x+4=0 |
| 7x=2 | x=-4 |
| x=\dfrac{2}{7} |
Therefore, the x-intercepts of y=7x^2+26x-8 are at \dfrac{2}{7} and at 4.
We can verify by creating the graph of y=7x^2+26x-8.
x-intercepts: x=\dfrac{3}{2} and x=-\dfrac{1}{3}
Video Demo of The Box Method
For our more visual learners, here’s a very quick video demo of factoring a trinomial using the box method:
Return to the Table of Contents
Solving Quadratic Equations by Factoring: World Problems
In real-life modeling situations, the x-intercepts can provide useful information on a graph! Quadratics can be used to model many things from arches used in construction to the path of a field goal kick.
Ball Thrown in the Air
Today, we will model the path of a ball that is thrown in the air as described below.
A ball is thrown from a height of 6 feet off the ground. The path of the ball is modeled by the equation y=-x^2+5x+6 where x is time measured in seconds and y is the height of the ball measured in feet.
How long does it take for the ball to land on the ground?
To start, we must remember that when something lands on the ground, the height is 0. In other words, when the ball lands on the ground, the value of y is 0. We can set the equation equal to 0
Now we have -x^2+5x+6=0. We know a=-1, b=5, and c=6.
Start With The Box Method
We can use the box method to factor this equation (read more about the box method here). We’ll put the term -x^2 in the top left space of the box and the term 6 in the bottom right space in the box.
| -x^2 | |
| 6 |
We now must determine the value of ac. In this case, ac = (-1)(6) = -6. We need the factor pair that multiplies to -6 and adds to 5, the value of b. So, we will list the factors where the number with the greater absolute value is positive because the factors must multiply to a negative number and add to a positive number.
| Relevant Factor Pairs of -6 | Sum of Factors |
|---|---|
| \color{blue}{6} and \color{blue}{-1} | \color{blue}{5} |
| -3 and 2 | -1 |
The factor pair 6 and -1 multiplies to -6 and adds to 5.
Now, we can fill in 6x and -x into the remaining places in the box.
| -x^2 | 6x |
| -x | 6 |
Now, we will take out the greatest common factor for each row and column. Let’s begin with the rows.
In row 1, the greatest common factor of -x^2 and 6x is x. In row 2, the greatest common factor of -x and 6 is 1.
| -x | 6 | |
| x | -x^2 | 6x |
| 1 | -x | 6 |
Therefore, we can rewrite the equation -x^2+5x+6=0 as:
(x+1)(-x+6)=0
We can solve for the roots, or x-intercepts, by setting each factor equal to 0.
| x+1=0 x=-1 | -x+6=0 -x=-6 x=6 |
This means the ball is on the ground when x=-1 and when x=6. We know x is the number of seconds that have passed. It is not logical to say the ball lands on the ground after -1 \text{ seconds} have passed. The model is only effective when the number of seconds is 0 or greater and when the number of feet off the ground is 0 or greater. We can thus discard illogical solutions like -1 seconds.
The solution of 6 \text{ seconds} is our final answer! The ball will land on the ground after 6 \text{ seconds} have passed.
Return to the Table of Contents
Determine a Quadratic Equation Given Its Roots
We have learned how to determine the roots, or x-intercepts, of a quadratic equation. Now, we will see how to work in reverse. If we know the roots of a quadratic equation, we are able to determine the equation! We can simply input the roots into the factored form of the equation.
Remember, factored form of a quadratic equation is y=(x-r_1)(x-r_2) where r_1 and r_2 are the roots, or x-intercepts, of the quadratic equation.
Using the roots -2 and 5
If the roots, or the x-intercepts, are -2 and 5, we can substitute -2 for r_1 and 5 for r_2.
This gives us the equation y=(x-(-2))(x-5).
We can simplify because we know that subtracting a negative is the same as adding a positive.
Our simplified equation is y=(x+2)(x-5).
To change this equation into standard form, we multiply the binomials. We can use double distribution.
y=(x+2)(x-5)=x^2-5x+2x-10=x^2-3x-10
Therefore, our equation is y=x^2-3x-10
Using the roots 3 and -\dfrac{1}{2}
Let’s show one more example using roots 3 and -\dfrac{1}{2}. We can substitute 3 for r_1 and -\dfrac{1}{2} for r_2.
This gives us the equation y=(x-3)(x-(-\dfrac{1}{2})).
We can simplify because we know that subtracting a negative is the same as adding a positive.
Our simplified equation is y=(x-3)(x+\dfrac{1}{2}).
To change this equation into standard form, we multiply the binomials. We can use double distribution.
(x-3)(x+\dfrac{1}{2})=x^2+\dfrac{1}{2}x-3x-\dfrac{3}{2}=x^2-\dfrac{5}{2}x-\dfrac{3}{2}
Therefore, our equation is y=x^2-\dfrac{5}{2}x-\dfrac{3}{2}.
Solving a Quadratic Equation Using Completing the Square
Solving by factoring is not the only method for solving a quadratic equation. All quadratic equations can be solved using completing the square, even equations that are not factorable.
In fact, this was the original method used, going back to the Babylonians according to Hackworth and Howland in the text Introductory College Mathematics: History of Real Numbers. For more on the history of solving quadratic equations, check out our article on the history of the quadratic formula.
Completing the square will also be useful when graphing other types of equations, such as circles and ellipses!
Example of Completing the Square
Let’s solve for the x-intercepts of the quadratic equation y=3x^2+9x-12, where a=3, b=9 and c=12. As before, we begin by setting y=0. Remember, we are solving for x-intercepts.
We have the equation 3x^2+9x-12=0. We will begin by dividing all terms by a, in this case 3.
Now, we have the equation x^2+3x-4=0.
We will subtract the value of c from both sides to keep the terms with a variable separate from terms without a variable. For us, this means subtracting -4, or adding 4, to each side.
This gives us the equation x^2+3x=4
Now, to complete the square, we must determine half of the coefficient in front of x and square that value. This will be the value to add to both sides of our equation. The coefficient in front of x is 3.
3 \times \dfrac{1}{2} = \dfrac{3}{2}
(\dfrac{3}{2})^2=\dfrac{9}{4}
Now, we will add \dfrac{9}{4} to both sides of the equation.
x^2+3x=4
x^2+3x+\dfrac{9}{4}=4+\dfrac{9}{4}
We will factor the left side because it is a perfect square. We will also simplify the right side of the equation. Remember that when we take the square root, we must include the positive and the negative solution.
x^2+3x+\dfrac{9}{4}=4+\dfrac{9}{4}
(x+\dfrac{3}{2})^2=-\dfrac{25}{4}
x+\dfrac{3}{2}=\pm \sqrt{\dfrac{25}{4}}
x+\dfrac{3}{2}=\pm {\dfrac{5}{2}}
Don’t Forget: Solve for x
To solve for x we will need to take the square root of both sides and isolate x by subtracting \dfrac{3}{2} from both sides.
x= \pm \dfrac{5}{2}-\dfrac{3}{2}
This gives us two solutions. We will separate the plus or minus and simplify each solution.
| x=\dfrac{5}{2}-\dfrac{3}{2} x=\dfrac{2}{2} x=1 | x=-\dfrac{5}{2}-\dfrac{3}{2} x=-\dfrac{8}{2} x=-4 |
The zeros, roots, or x-intercepts of the equation y=3x^2+9x-12 are 1 and -4.
This method can take a bit longer, which is why factoring is such a useful strategy! Completing the square is still a useful skill that serves as the foundation for graphing and solving other types of equations. It is good to have a variety of tools on our tool belt so that we are able to efficiently solve many types of problems.
Return to the Table of Contents
Practice with Factoring Quadratics
For practice questions focused on solving quadratic equations by factoring, explore Albert’s Algebra 1 practice course! All Albert questions include explanations of solutions and tips for avoiding common mistakes.
If you’re more of a visual learner, you can watch this short video demonstrating how to solve quadratic equations using factoring.
Additionally, licensed Albert teachers can assign students this short Algebra 1 Topic Quiz that focuses on solving quadratic equations through factoring and graphing.
Finally, check out our other detailed Algebra 1 review guides to learn more about quadratics.
