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

Before we start looking at how to solve a quadratic equation by factoring, let us first consider the graph of the quadratic equation .

When we talk about solving a quadratic equation, we are talking about identifying the roots of the quadratic equation which are the points where the graph crosses the -axis (the -intercepts), that is, the point such that . We can see from the graph that the roots of are and . We leave this for a second and now look at factoring the quadratic. We first need to identify the factor pairs of 12; we have

We can then see from these factor pairs that and, therefore, the quadratic can be factored to

At this point, you may notice that the numbers in the parentheses are the same as the roots of the quadratic but the signs are opposite. Let us look at this a little more closely. As previously noted, we can find the roots of our quadratic by finding the values of that give an output of zero, that is, solve the equation

We have shown that the quadratic can be written in factored form, thus rewritten as

If we think of these two binomials as two numbers multiplied together, the only way that we can get an answer of zero is if one of the numbers is zero. Therefore, our solutions can be found by solving each of the following equations:

If we subtract 6 from each side of the first equation, we get and if we add 2 to each side of the second equation we find that (which are the roots as identified from our graph).

One point here that is worth noting is that, for monic quadratics (where the leading coefficient is one), the roots will be the same as the numbers in the factored form but with opposite signs. This is not true of nonmonic quadratics however.

There are generally three types of core questions when solving quadratic equations by factoring; the first involves equations like where the expressions factor into a single bracket; the second involves equations like those we have just looked at; that is, where the coefficient of the leading terms is one; and the third involves equations like where the quadratics are nonmonic. We may also encounter questions where the first step is to rearrange the equation to get it in a standard form that we know how to solve. We will have a look at each of the three types of questions now.

### Example 1: Finding the Roots of a Quadratic Equation in the Form ππ₯^{2} + ππ₯ = 0

- Factor the equation .
- At which values of does the graph of cross the -axis?

### Answer

With this question, solving the first part points us in the right direction for solving the second part. To factor the expression for part one, we need to identify the highest common divisor for each of the two terms in the expression. The number 3 is the greatest number that divides each of the two terms and is the greatest variable. Our greatest common divisor is therefore . If we then divide each of the terms by this divisor, we get and 3, which means our expression can be factored as follows:

We can always check this by expanding the expression. That is, , which is correct.

To solve the second part, we need to set our factored expression equal to zero and then solve the following equation:

The left-hand side of this equation will only be equal to zero if either or is equal to zero. Therefore, to solve the equation, we can solve each of the following equations:

If we divide each side of the first equation by 3, we find that , and if we subtract 3 from each side of the second equation and then divide by 2, we get . Therefore, the solutions to our quadratic are and .

### Example 2: Finding the Roots of a Quadratic Equation in the Form π₯^{2} + ππ₯ + π = 0

Solve by factoring.

### Answer

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 need to consider the factor pairs of the constant term 4. We have

We then need to use one of these factor pairs to make the coefficient of , which we can do here using 2, 2; that is,

This means that our expression factors to

This product of binomials can only be zero if one of the binomials equals zero. In this case, the binomials are equal so we only have one solution which can be found by solving the equation

If we add 2 to each side, we find that

### Example 3: Finding the Roots of a Quadratic Equation in the Form ππ₯^{2} + ππ₯ + π = 0

Solve the equation by factoring.

### Answer

Here, we have an equation that contains a nonmonic quadratic, that is, a quadratic where the leading term is not equal to one. To factor this expression, we may notice that it is a perfect square as and are both square numbers, which means it factors to .

If we do not immediately notice this, we can try using trial and error, or we can take a more systematic approach. We can multiply the and , and then rewrite in terms of a factor pair of . If we write down the factor pairs of 225, we get

We can then rewrite our equation as

We then factor the first two terms and the last two terms to give us

If we factor out the binomial , we get

This product of binomials can only be zero if one of the binomials equals zero. In this case, the binomials are equal so we only have one solution which can be found by solving the equation

If we subtract 5 from each side and divide by 3, we find that

As seen in the previous example, it is always worth keeping an eye out for special types of quadratics to aid the factoring process. In the previous example, we saw the factoring of a perfect square, but we may also see an example of a difference of two squares, that is, expressions in the form

Let us now have a look at an example containing a difference of two squares.

### Example 4: Finding the Roots of a Quadratic Equation in the Form π^{2} β π^{2} = 0

At which values of does the graph of cross the -axis?

### Answer

Here, we have been asked to find the points at which the graph crosses the -axis. The points are also known as the roots of the equation or, indeed, the solutions to the equation . To solve this, we first need to factor the left-hand side. The quadratic expression is actually a difference of two squares, which means that it factors as follows:

This product of binomials can only be zero if one of the binomials equals zero. Therefore, we can find the solutions to the equation by solving each of the following equations:

If we add to each side of the first equation, we find that and if we subtract from each side of the second equation we find that . These are the roots of the quadratic and, indeed, the values of the points where the graph crosses the -axis.

To finish, let us look at one final example where we can take a slightly different approach to find the solution using the information given in the question.

### Example 5: Finding a Root of a Quadratic Equation Given Its Other Root

Given that is a root of the equation , what is the other root?

### Answer

We are told that is a root of the equation, which means that our quadratic is zero when . This means that is a factor of the equation. This means that there will be another factor such that

From here, by comparing coefficients, we can see that which gives . This means that we can rewrite our original equation as

We already know that one solution is , and we can find the second solution by solving the equation

By subtracting 4 from each side and then dividing through by 5, we find that

### Key Points

- Determine if the quadratic factors into a product of two binomials or a product of a monomial and a binomial. Generally, if the quadratic is in the form where , , and are nonzero, the quadratic will factor into two binomials. If is zero, then the quadratic will factor into a monomial and a binomial.
- For a quadratic in the form , where and and are nonzero, the quadratic factors into the form , where .
- For a quadratic in the form , where , and , and are nonzero, this can be factored by finding a factor pair of , say , , such that . At this point, we can rewrite the quadratic as and then factor each of the expressions and .