Lesson Explainer: Solving Quadratic Equations: Factoring Mathematics • 9th Grade

In this explainer, we will learn how to solve quadratic equations by factoring.

Let us begin by recalling the definition of a quadratic equation.

Definition: Quadratic Equation

A quadratic equation is any equation that can be expressed in the standard form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are real constants and π‘Žβ‰ 0.

We note that a quadratic equation can have zero, one, or two solutions (i.e., values of π‘₯ that satisfy the equation). To find the solutions to a quadratic equation, there are multiple possibilities available to us, but in this explainer we will focus on the method of factoring.

Recall that to factor something means to split it into separate factors (often in parentheses) that multiply to make the whole thing. For instance, the expression π‘₯βˆ’5π‘₯ can be factored by noting that π‘₯ is a common factor of both terms, meaning we can write it as π‘₯(π‘₯βˆ’5).

Suppose that we wanted to find the solutions to the quadratic equation π‘₯βˆ’5π‘₯=0. By considering the factored form of the left-hand side, it becomes easy to identify what the solutions must be. That is, we have

As is highlighted above, if the product of two (or more) factors is equal to zero, then at least one of the individual factors must itself be equal to zero. Since there are two factors, we therefore have two possibilities: π‘₯=0π‘₯βˆ’5=0.or

By adding 5 to both sides of the second equation, we get π‘₯=0π‘₯=5.or

Therefore, there are two solutions (also known as roots) for the equation, π‘₯=0 and π‘₯=5. If required, these numbers can also be written as a solution set as {0,5}.

Let us see another example of factoring a quadratic in such a way to solve it.

Example 1: Finding the Solution Set of Quadratic Equations by Factoring

Find the solution set of π‘₯+12π‘₯=0 in ℝ.


We have been asked to find the solution set of this equation in ℝ (i.e., the real numbers), which we can do by factoring the left-hand side. To do this, we begin by recognizing that both terms have a common factor of π‘₯. Thus, we can factor them to get π‘₯(π‘₯+12)=0.

Now, we can find the solutions to this equations by setting each factor equal to 0 in turn. So we have π‘₯=0π‘₯+12=0.or

For the second equation, we can subtract 12 from both sides to put it in terms of π‘₯: π‘₯=0π‘₯=βˆ’12.or

Thus, the solution set is {βˆ’12,0}.

In the previous example, factoring the quadratic equation was fairly straightforward because both terms shared a common factor of π‘₯, but in general, we will have to do more work to obtain the factored form. For instance, suppose we have π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where all of π‘Ž, 𝑏, and 𝑐 are nonzero. To approach factoring a quadratic of this form, recall that we have a variety of methods to help us, including

  • factoring by inspection,
  • recognizing a quadratic as a perfect square,
  • recognizing a quadratic as the difference of two squares,
  • factoring by grouping.

Suppose that by applying one of the above techniques we are able to factor the quadratic as follows: (𝑝π‘₯+π‘ž)(π‘Ÿπ‘₯+𝑠)=0, where 𝑝 and π‘Ÿ are nonzero. Then, much like in the previous example, we can determine the solutions by setting each factor equal to zero in turn. This gives us 𝑝π‘₯+π‘ž=0π‘Ÿπ‘₯+𝑠=0.or

We can then solve each equation separately to get 𝑝π‘₯+π‘ž=0π‘Ÿπ‘₯+𝑠=0𝑝π‘₯=βˆ’π‘žπ‘Ÿπ‘₯=βˆ’π‘ π‘₯=βˆ’π‘žπ‘,π‘₯=βˆ’π‘ π‘Ÿ.

Thus, by factoring a quadratic, it becomes easy to identify the solutions.

Let us consider an example of a quadratic that must be factored into two parentheses to solve it.

Example 2: Finding the Roots of a Quadratic Equation in the Form π‘₯2 + 𝑏π‘₯ + 𝑐 = 0

Solve π‘₯βˆ’4π‘₯+4=0 by factoring.


We are told in the question that we need to solve this quadratic equation by factoring, so our first step is to factor the left-hand side of the equation. To do this, we can either recognize that the form the quadratic takes is a perfect square or we can apply a more universal approach and use inspection. For the purposes of generality, we will do the latter.

To factor by inspection, recall that we can take factor pairs that multiply to make the last term, and find which ones sum to make the middle term. Since the last term is 4, the distinct factor pairs are 1Γ—4=4⟹14,2Γ—2=4⟹22.andand

Next, we consider the middle term, βˆ’4π‘₯, which has a coefficient of βˆ’4. While we cannot add 1 and 4 (or βˆ’1 and βˆ’4) to make βˆ’4, using two lots of βˆ’2, we have βˆ’2βˆ’2=βˆ’4.

Thus, we know that βˆ’2 appears twice in the factors, meaning we can factor the quadratic as follows: π‘₯βˆ’4π‘₯+4=(π‘₯βˆ’2)(π‘₯βˆ’2)=(π‘₯βˆ’2).

This can be verified by expanding the parentheses. So, our equation can be written as (π‘₯βˆ’2)=0.

We can solve this equation, since we know it can only be zero when π‘₯βˆ’2=0. By adding 2 to each side, we get the solution, π‘₯=2.

In the previous example, we had to deal with a quadratic equation where the leading term (i.e., the π‘₯ term) had a coefficient of 1, but sometimes we will need to be able to factor quadratics where this is not the case.

In the next example, we will look toward factoring and solving a nonmonic quadratic equation.

Example 3: Finding the Roots of a Quadratic Equation in the Form π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0

Find the solution set of 6π‘₯βˆ’13π‘₯+6=0 in ℝ.


To find the solutions to this quadratic, we can factor the left-hand side using inspection. To do this, since it has been given in the standard form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, we can begin by multiplying π‘Ž and 𝑐 (i.e., 6Γ—6=36), listing the factors of this number, and identifying the factor pair whose sum is 𝑏 (i.e., βˆ’13):

So, let us rewrite the expression using these values: 6π‘₯βˆ’4π‘₯βˆ’9π‘₯+6=0.

Now, we can factor this by grouping the first two and last two terms: 6π‘₯βˆ’4π‘₯ο‡Œο†²ο‡ο†²ο‡Ž+βˆ’9π‘₯+6ο‡Œο†²ο‡ο†²ο‡Ž=2π‘₯(3π‘₯βˆ’2)βˆ’3(3π‘₯βˆ’2)=(2π‘₯βˆ’3)(3π‘₯βˆ’2).οŠ¨οŠ¨ο—οŠ©FactorofFactorof

Finally, now that the quadratic has been factored, we can solve the equation by setting each factor equal to 0. That is, we have (2π‘₯βˆ’3)(3π‘₯βˆ’2)=02π‘₯βˆ’3=03π‘₯βˆ’2=02π‘₯=33π‘₯=2π‘₯=32π‘₯=23.ororor

Written as a solution set, this is 23,32.

So far, we have only seen quadratic equations given in standard form (i.e., π‘Žπ‘₯+𝑏π‘₯+𝑐=0), but sometimes they may be in a different form despite still essentially being quadratics. As one example, we could have something like (π‘₯+4)=9.

Although we can see that the left-hand side is in a factored form, it does not help us to immediately solve the equation. Instead, we have to manipulate it into a form that we can find the solutions to. In this case, if we take the square root of both sides, we get π‘₯+4=Β±3.

Note that the right-hand side can be positive or negative, since both 3 and βˆ’3 are square roots of 9. This results in two equations: π‘₯+4=βˆ’3π‘₯+4=3,π‘₯=βˆ’7π‘₯=1.oror

Thus, the solutions are π‘₯=βˆ’7 and 1. While we did not have to do too much extra work in this case, in general, we might have to do additional computations. In particular, an effective approach to these sorts of problems is to first rearrange the equation into standard form before we try to solve it. Let us see an example of this.

Example 4: Solving a Quadratic Equation Not Given in Standard Form

Find the solution set of ο€Ήπ‘₯βˆ’3π‘₯=βˆ’8 in ℝ.


As this equation is not in the standard form for a quadratic equation (i.e., π‘Žπ‘₯+𝑏π‘₯+𝑐=0), we will have to rearrange it into that form before we can solve it.

To begin doing this, we should eliminate the cube on the left-hand side, which we can do by taking the cube root of both sides. This gives us οŽ’οŽ’ο„(π‘₯βˆ’3π‘₯)=βˆšβˆ’8π‘₯βˆ’3π‘₯=βˆ’2.

Then, by adding two to both sides, we can convert it into standard form: π‘₯βˆ’3π‘₯+2=0.

Now, we can solve this equation by factoring. We can do this by inspection by considering a factor pair that multiply to make 2 and add to get βˆ’3. This can be done with βˆ’2 and βˆ’1. So, we have π‘₯βˆ’2π‘₯βˆ’π‘₯+2=0(π‘₯βˆ’2)(π‘₯βˆ’1)=0.

Now, we can solve this by setting the factors to be zero. This gives us two linear equations to solve: π‘₯βˆ’2=0π‘₯βˆ’1=0,π‘₯=2π‘₯=1.oror

As a solution set, this is {1,2}.

As a side comment to the kind of rearrangement demonstrated in the previous example, we should be careful when squaring or taking square roots of equations to rearrange them. Since we were taking a cube root, we did not have to do anything in particular (since cube roots are valid for any real number), but say for example we had √4π‘₯+3=2π‘₯βˆ’6.

If we squared both sides and then factored the resulting quadratic equation (although we will not cover the steps here), we would eventually get (2π‘₯βˆ’3)(2π‘₯βˆ’11)=0.

Although this would normally have two solutions, which are π‘₯=32 and π‘₯=112, we note that π‘₯=32 would result in the right-hand side of the original equation being negative, while the left-hand side has a square root that can only be nonnegative. Therefore, there would only be one correct solution. We should always keep in mind that square roots can potentially lead to only certain solutions being valid.

For our final example, let us consider a geometric problem that can be solved using the factoring methods we have learned so far.

Example 5: Finding an Unknown in the Dimensions of a Shape by Solving a Quadratic Equation

Given that the area of the shape below is 7 units, what is the value of π‘₯?


To solve this problem, we must recall the formula for the area 𝐴 of a trapezoid: 𝐴=π‘Ž+𝑏2β„Ž, where β„Ž is the height, and π‘Ž and 𝑏 are the opposite parallel sides. If we substitute π‘Ž=π‘₯+2, 𝑏=π‘₯+6, β„Ž=π‘₯βˆ’2, and the given area 𝐴=7 into the formula, we get 7=(π‘₯+2)+(π‘₯+6)2(π‘₯βˆ’2)=2π‘₯+82(π‘₯βˆ’2)=(π‘₯+4)(π‘₯βˆ’2).

While this equation is in a factored form, we cannot yet solve it for π‘₯ as there is a constant value of 7 on the left-hand side. Thus, let us take everything to the same side: (π‘₯+4)(π‘₯βˆ’2)βˆ’7=0.

To solve this, we will first need to expand the parentheses so that we can get it into the standard form π‘Žπ‘₯+𝑏π‘₯+𝑐=0. Doing this, we get π‘₯+4π‘₯βˆ’2π‘₯βˆ’8βˆ’7=0π‘₯+2π‘₯βˆ’15=0.

Now, we factor it again. We do this by finding two numbers that multiply to make 𝑐, which is βˆ’15, and add to get 𝑏, 2:

So, using βˆ’3 and 5, we can factor the quadratic expression to get π‘₯+2π‘₯βˆ’15=(π‘₯βˆ’3)(π‘₯+5).

Now, setting this to zero, we can find the solutions by setting each factor to zero as follows: (π‘₯βˆ’3)(π‘₯+5)=0π‘₯βˆ’3=0π‘₯+5=0π‘₯=3π‘₯=βˆ’5.oror

We have two possible solutions, but since all of the lengths of the shape must be positive, we can see that π‘₯=βˆ’5 is not an option. Thus, π‘₯ must be 3.

Let us finish by summarizing the main things we have learned in this explainer.

Key Points

  • The solutions to a factored quadratic equation that is equal to zero can be obtained by setting each factor equal to zero.
  • If the quadratic equation is in standard form, we can factor it using methods we already know for factoring quadratics so that it can be solved.
  • If the quadratic equation is not in standard form, we can begin by rearranging it into this form (sometimes squaring and taking square roots may be necessary) before continuing with the procedure for a quadratic in standard form.

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