Lesson Explainer: Solving Quadratic Equations: Factoring Mathematics

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 𝑦=𝑥+4𝑥12.

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 𝑥-axis (the 𝑥-intercepts), that is, the point 𝑥 such that 𝑥+4𝑥12=0.

We can see from the graph that the roots of 𝑦=𝑥+4𝑥12 are 𝑥=6 and 𝑥=2. 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 1,122,63,4.

We can then see from these factor pairs that +62=4 and, therefore, the quadratic can be factored to (𝑥+6)(𝑥2).

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 𝑥 that give an output of zero, that is, solve the equation 𝑥+4𝑥12=0.

We recall the method of factoring quadratics. When factoring, we are looking for a pair of numbers whose product is 12 and whose sum is 4. Noticing that 6×(2)=12,6+(2)=4,and we obtained the factored form of the quadratic equation above as (𝑥+6)(𝑥2)=0.

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: 𝑥+6=0𝑥2=0.and

If we subtract 6 from each side of the first equation, we get 𝑥=6 and if we add 2 to each side of the second equation we find that 𝑥=2 (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 4𝑥+8𝑥=0, where we can factor out the common factor 4𝑥 from the expression on the left hand side of the equation to write 4𝑥(𝑥+2)=0.

The second involves quadratic expressions with a nonzero constant term the leading coefficient 1, such as 𝑥+5𝑥+6=0.

The third involves quadratic expressions with a nonzero constant term and a leading coefficient different from 1, such as 6𝑥5𝑥4=0.

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. Let us consider each of the three types of quadratic equations. Let us begin by considering the first type of quadratic equations which involve a quadratic expression with a common factor.

Example 1: Finding the Roots of a Quadratic Equation in the Form 𝑎𝑥2 + 𝑏𝑥 = 0

  1. Factor the equation 𝑦=6𝑥+9𝑥.
  2. At which values of 𝑥 does the graph of 𝑦=6𝑥+9𝑥 cross the 𝑥-axis?

Answer

With this question, solving the first part points us in the right direction for solving the second part.

Part 1

To factor the expression, 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 both 6𝑥 and 9𝑥 share the term 𝑥. Our greatest common divisor is therefore 3𝑥. If we then divide each of the terms by this divisor, we get 2𝑥 and 3, which means our expression can be factored as follows: 3𝑥(2𝑥+3).

We can check this by expanding the expression. That is, 3𝑥×2𝑥+3𝑥×3=6𝑥+9𝑥, which is correct.

Part 2

We need to set our factored expression equal to zero and then solve the following equation: 3𝑥(2𝑥+3)=0.

The left-hand side of this equation will only be equal to zero if either 3𝑥 or 2𝑥+3 is equal to zero. Therefore, to solve the equation, we can solve each of the following equations: 3𝑥=02𝑥+3=0.and

If we divide each side of the first equation by 3, we find that 𝑥=0, and if we subtract 3 from each side of the second equation and then divide by 2, we get 𝑥=32. Therefore, the solutions to our quadratic are 𝑥=0 and 𝑥=32.

In the next example, we consider the second type of quadratic equations where the quadratic expression involves a nonzero constant term and the leading coefficient is equal to 1.

Example 2: Finding the Roots of a Quadratic Equation in the Form 𝑥2 + 𝑏𝑥 + 𝑐 = 0

Solve 𝑥4𝑥+4=0 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 four distinct pairs 1,42,2.

We then need to use one of these factor pairs to make the coefficient of 𝑥, which we can do here using 2, 2; that is, 22=4.

This means that our expression factors to (𝑥2)(𝑥2)=0.

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 𝑥2=0.

If we add 2 to each side, we find that 𝑥=2.

Let us now consider the last type of quadratic equations where the quadratic expression contains a nonzero constant and the leading coefficient is not equal to 1.

Example 3: Finding the Roots of a Quadratic Equation in the Form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

Solve the equation 9𝑥+30𝑥+25=0 by factoring.

Answer

Here, we have an equation that contains a nonmonic quadratic, that is, a quadratic where the coefficient of the leading term is not equal to one. To factor this expression, we may notice that it is a perfect square as 𝑎 and 𝑐 are both square numbers, which means it factors to (3𝑥+5).

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 𝑎=9 and 𝑐=25, and then rewrite 𝑏 in terms of a factor pair of 𝑎𝑐. If we write down the factor pairs of 225, we get 1,2253,755,459,2515,15.

We can then rewrite our equation as 9𝑥+15𝑥+15𝑥+25=0.

We then factor the first two terms and the last two terms to give us 3𝑥(3𝑥+5)+5(3𝑥+5)=0.

If we factor out the binomial (3𝑥+5), we get (3𝑥+5)(3𝑥+5)=0.

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 3𝑥+5=0.

If we subtract 5 from each side and divide by 3, we find that 𝑥=53.

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 𝑎𝑏=(𝑎+𝑏)(𝑎𝑏).

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 𝑎2 ‒ 𝑏2 = 0

At which values of 𝑥 does the graph of 𝑦=𝑥7 cross the 𝑥-axis?

Answer

Here, we have been asked to find the points at which the graph 𝑦=𝑥7 crosses the 𝑥-axis. The points are also known as the roots of the equation or, indeed, the solutions to the equation 𝑥7=0. 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: 𝑥7𝑥+7=0.

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: 𝑥7=0𝑥+7=0.and

If we add 7 to each side of the first equation, we find that 𝑥=7 and if we subtract 7 from each side of the second equation we find that 𝑥=7. These are the roots of the quadratic and, indeed, the 𝑥 values of the points where the graph crosses the 𝑥-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 15 is a root of the equation 5𝑥+79𝑥+60=0, what is the other root?

Answer

We are told that 15 is a root of the equation, which means that our quadratic is zero when 𝑥=15. This means that 𝑥+15 is a factor of the equation. This means that there will be another factor 𝑎𝑥+𝑏 such that (𝑥+15)(𝑎𝑥+𝑏)=5𝑥+79𝑥+60.

Let us expand through the parenthesis on the left hand side of the equation above. (𝑥+15)(𝑎𝑥+𝑏)=𝑎𝑥+15𝑎𝑥+𝑏𝑥+15𝑏=𝑎𝑥+(15𝑎+𝑏)𝑥+15𝑏.

Comparing the coefficient of 𝑥 and the constant term in the right hand side of the original equation, we obtain 𝑎=515𝑏=60,and which gives 𝑏=4. This means that we can rewrite our original equation as (𝑥+15)(5𝑥+4)=0.

We already know that one solution is 15, and we can find the second solution by solving the equation 5𝑥+4=0.

By subtracting 4 from each side and then dividing through by 5, we find that 𝑥=45.

Let us recap a few important concepts from this explainer.

Key Points

  • Solutions of a factored quadratic equation is obtained by setting each factor (binomial) equal to zero. Quadratic expressions of different types can be factored as listed below.
    • For a quadratic in the form 𝑎𝑥+𝑏𝑥, we can factor the common factor which is gcd (𝑎,𝑏)𝑥.
    • For a quadratic in the form 𝑎𝑥+𝑏𝑥+𝑐, where 𝑎=1 and 𝑏 and 𝑐 are nonzero, if the quadratic has real-valued roots, the quadratic factors into the form (𝑥+𝑘)(𝑥+𝑙), where 𝑘𝑙=𝑐 and 𝑘+𝑙=𝑏.
    • For a quadratic in the form 𝑎𝑥+𝑏𝑥+𝑐, where 𝑎1, and 𝑎, 𝑏 and 𝑐 are nonzero, if the quadratic has real-valued roots, this can be factored by finding a factor pair of 𝑎𝑐, say 𝑘, 𝑙, such that 𝑏=𝑘+𝑙. At this point, we can rewrite the quadratic as 𝑎𝑥+𝑘𝑥+𝑙𝑥+𝑐 and then factor each of the expressions 𝑎𝑥+𝑘𝑥 and 𝑙𝑥+𝑐.
    • For a quadratic in the form 𝑥𝑐 where 𝑐 is positive, the quadratic factors into 𝑥𝑐𝑥+𝑐.

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