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Lesson Explainer: Factoring Trinomials Mathematics

In this explainer, we will learn how to factor trinomials into a product of two binomials.

We begin by recalling the definitions of monomials, binomials, and trinomials.

Definitions: Monomial, Binomial, and Trinomial

A monomial is a product of numbers and powers of variables.

A binomial expression is the sum or difference of two monomials.

A trinomial expression is the sum or difference of three monomials.

An example of a monomial expression is βˆ’5π‘₯π‘¦οŠ¨.

An example of a binomial expression is 3π‘₯+7.

An example of a trinomial expression is 2π‘Žβˆ’2π‘Žπ‘+3π‘οŠ¨.

When we list the factors of a number, we can write the number as a product of its factors. For example, we may write 20=2Γ—10. The same principle is true when we factor algebraic expressions; the focus of this explainer is writing trinomials as the product of two binomial factors.

In general, when we multiply two binomials, we initially obtain four terms, created by multiplying each term in one binomial by each term in the other. If the two binomials have the same algebraic structure, then we are be able to combine one pair of like terms, leading to a trinomial. Each of the examples we consider here are of this type.

We first demonstrate the process of factoring an expression with a common binomial term that is already in a partially factored form.

Example 1: Factoring an Expression with a Common Binomial Term

Fully factor the expression π‘₯(π‘₯+3)+2(π‘₯+3).

Answer

Upon inspection, we observe that the two parts of this expression share a common binomial factor of (π‘₯+3): π‘₯(π‘₯+3)+2(π‘₯+3).

We can, therefore, factor by this shared binomial. In the first part of the expression, this binomial is multiplied by π‘₯, and in the second part, it is multiplied by 2. Hence, overall, it is multiplied by (π‘₯+2), and so π‘₯(π‘₯+3)+2(π‘₯+3)=(π‘₯+3)(π‘₯+2).

This expression cannot be factored any further, as the two terms in each binomial do not share any common factors other than 1.

We now consider how to factor a quadratic expression of the form π‘₯+𝑏π‘₯+π‘οŠ¨ into the product of two binomials. Factoring is the reverse process of distributing parentheses, or expanding brackets. Consider the expansion of the product of the binomials (π‘₯+3) and (π‘₯+5). We begin by distributing the first set of parentheses over the second: (π‘₯+3)(π‘₯+5)=π‘₯(π‘₯+5)+3(π‘₯+5).

Distributing each set of parentheses and collecting like terms gives π‘₯(π‘₯+5)+3(π‘₯+5)=π‘₯+5π‘₯+3π‘₯+15=π‘₯+8π‘₯+15.

We can observe that the constant term in the trinomial expression is the product of the constant terms in the two binomials (15=3Γ—5). The coefficient of π‘₯ in the trinomial is the sum of the constant terms in the two binomials (8=3+5) and in the penultimate line of our solution, the π‘₯-term is written as the sum of two terms with these coefficients (5π‘₯+3π‘₯). This suggests a process we can follow to work in the opposite direction and factor an expanded quadratic of the form π‘₯+𝑏π‘₯+π‘οŠ¨ into the product of two binomials. It is important to note that not all quadratics of this form can be factored, so the process we outline here is only applicable to those that are factorable.

How To: Factoring a Quadratic of the Form π‘₯2 + 𝑏π‘₯ + 𝑐 into the Product of Two Binomials

  • List the factor pairs of the constant term 𝑐.
  • If 𝑐 is positive, the two numbers will have the same sign, whereas if 𝑐 is negative, the two numbers will have opposite signs.
  • Look for a factor pair that, with the correct combination of signs, sum to the coefficient of π‘₯(𝑏).
  • Rewrite the middle term in the trinomial (𝑏π‘₯) as the sum of two terms with coefficients equal to the factor pair found.
  • Separate the new four-term expression into two binomials and factor each.
  • Look for a shared binomial factor to factor the entire expression by.

We will demonstrate the process of factoring a quadratic of the form π‘₯+𝑏π‘₯+π‘οŠ¨ in our next example.

Example 2: Factoring a Monic Quadratic into a Product of Binomials

Factor π‘₯βˆ’8π‘₯βˆ’20.

Answer

To factor this quadratic expression, we wish to write it as the product of two binomials. We will do this by first rewriting the middle term as the sum of two terms with coefficients whose sum is the coefficient of π‘₯(βˆ’8) and whose product is the constant term (βˆ’20). We first consider the factor pairs of 20.

As the product of the two numbers must be βˆ’20, the two numbers must have different signs. If we choose the second factor pair of 2 and 10 and choose the 2 to be positive and the 10 to be negative, then the sum of these two numbers is 2+(βˆ’10)=βˆ’8 as required. We then rewrite the trinomial with the middle term expressed as the sum of two terms with coefficients of 2 and βˆ’10: π‘₯βˆ’8π‘₯βˆ’20=π‘₯+2π‘₯βˆ’10π‘₯βˆ’20.

Separating this four-term expression into two binomials and factoring each gives

Finally, we factor the entire expression by the shared binomial factor of (π‘₯+2) to give π‘₯βˆ’8π‘₯βˆ’20=(π‘₯+2)(π‘₯βˆ’10).

We could simplify the process used in the previous example by considering what the structure of the factored form of the quadratic would be. As the coefficient of π‘₯ is 1, the first term in each binomial must be π‘₯. Hence, the factored form of this quadratic is (π‘₯+π‘š)(π‘₯+𝑛) for constants π‘š and 𝑛 that we need to determine. Distributing the parentheses, we obtain (π‘₯+π‘š)(π‘₯+𝑛)=π‘₯(π‘₯+𝑛)+π‘š(π‘₯+𝑛)=π‘₯+𝑛π‘₯+π‘šπ‘₯+π‘šπ‘›=π‘₯+(π‘š+𝑛)π‘₯+π‘šπ‘›.

Comparing this to the original quadratic reveals that the numbers π‘š and 𝑛 have two properties:

  • Their sum must be equal to the coefficient of π‘₯(βˆ’8).
  • Their product must be equal to the constant term (βˆ’20).

These were the same conditions placed on the two numbers we used to rewrite the middle term of the quadratic as a sum. Hence, once we had identified that the pair of numbers satisfying these conditions were 2 and βˆ’10, we could simply replace π‘š and 𝑛 with these two values to complete each binomial, once again leading to (π‘₯+2)(π‘₯βˆ’10). This is a slightly more efficient method that can be useful when factoring quadratics of this form.

As we become more familiar with factoring quadratics into two binomials, we may be able to identify the two numbers with a given sum and product mentally. Whilst we are still learning this skill, however, we may find it helpful to list all possible combinations of each factor pair and signs in a table to help us identify the two numbers with the correct sum. FactorsofSumβˆ’20βˆ’120191βˆ’20βˆ’19βˆ’21082βˆ’10βˆ’8βˆ’4514βˆ’5βˆ’1

We have now seen how to factor quadratics of the form π‘₯+𝑏π‘₯+π‘οŠ¨ into the product of two binomials. Such quadratics, in which the coefficient of π‘₯ is equal to 1, are known as monic quadratics. We will now consider the more general case of how to factor a nonmonic quadratic of the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, where π‘Ž is not equal to 0 or 1 (or βˆ’1).

We will demonstrate this process through an example. Consider the product of the binomials (3π‘₯βˆ’2) and (2π‘₯+5). Using the distributive property, we obtain (3π‘₯βˆ’2)(2π‘₯+5)=3π‘₯(2π‘₯+5)βˆ’2(2π‘₯+5)=6π‘₯+15π‘₯βˆ’4π‘₯βˆ’10=6π‘₯+11π‘₯βˆ’10.

Now we consider the reverse of this: expressing the trinomial 6π‘₯+11π‘₯βˆ’10 as the product of two binomials. We observe that in the first line of our working, the two parts of the expression had a shared binomial factor of (2π‘₯+5). Upon distributing the parentheses, we obtained four terms, which simplified to three terms by combining like terms.

To work in the other direction, we first need to rewrite a trinomial as an expression involving four terms so that we can then separate the resulting expression into one that contains two binomials and factor each separately. The process for doing so is as follows.

How To: Factoring a Quadratic of the Form π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 into the Product of Two Binomials

  • List the factor pairs of π‘Žπ‘.
  • If π‘Žπ‘ is positive, the two numbers will have the same sign, whereas if π‘Žπ‘ is negative, the two numbers will have opposite signs.
  • Look for a factor pair that, with the correct combination of signs, sum to the coefficient of π‘₯(𝑏).
  • Rewrite the middle term in the trinomial (𝑏π‘₯) as the sum of terms with coefficients equal to the factor pair found.
  • Separate the new four-term expression into two binomials and factor each.
  • Look for a shared binomial factor to factor the entire expression by.

Taking the above example in reverse, we look for two numbers whose sum is the coefficient of π‘₯(11) and whose product is equal to the product of the coefficient of π‘₯ and the constant term (6Γ—βˆ’10=βˆ’60). These two numbers are 15 and βˆ’4. We then rewrite the trinomial as a four-term expression, separating the π‘₯-term into the sum of two terms with these coefficients: 6π‘₯+11π‘₯βˆ’10=6π‘₯+15π‘₯βˆ’4π‘₯βˆ’10.

We then split this new expression into two binomials and factor each binomial separately:

This reveals a common binomial factor of (2π‘₯+5), which can subsequently be factored to give our solution: 3π‘₯(2π‘₯+5)βˆ’2(2π‘₯+5)=(2π‘₯+5)(3π‘₯βˆ’2).

If after factoring each binomial we do not find a common binomial factor, this suggests that either we have made an error and should check our working carefully or the trinomial is not factorable into the product of two binomials.

We may be concerned that the order in which we rewrite the π‘₯-term as the sum of two terms matters. Suppose instead we had written the terms in the opposite order: 6π‘₯+11π‘₯βˆ’10=6π‘₯βˆ’4π‘₯+15π‘₯βˆ’10.

Proceeding as before, we find that

This time, the shared binomial factor was the other factor of (3π‘₯βˆ’2), and the binomial factors in our solution are written in the opposite order. However, as multiplication is commutative, our answer is equivalent, and so although the working will look a little different, the order in which we write the sum does not matter.

The method for factoring a monic quadratic that we met earlier is, in fact, a special case of this method, in which the product of π‘Ž and 𝑐 is equal to 𝑐 (because π‘Ž is equal to 1).

We will now consider an example in which we apply this process to factor a single variable trinomial into the product of three linear factors.

Example 3: Factorizing Trinomials by Taking Out the Greatest Common Factor

Factorize fully 6π‘₯βˆ’17π‘₯+12π‘₯.

Answer

We begin by observing that the three terms in this trinomial share a common factor of π‘₯. Hence, we can factor by π‘₯ to give 6π‘₯βˆ’17π‘₯+12π‘₯=π‘₯ο€Ή6π‘₯βˆ’17π‘₯+12.

We have now written the cubic expression as the product of a linear term (π‘₯) and a nonmonic quadratic ο€Ή6π‘₯βˆ’17π‘₯+12ο…οŠ¨. To fully factor this trinomial, we need to factor the quadratic.

We begin by finding two numbers whose sum is equal to the coefficient of π‘₯(βˆ’17) and whose product is equal to the product of the coefficient of π‘₯ and the constant term (6Γ—12=72). These two numbers are βˆ’8 and βˆ’9. We now separate the π‘₯-term into the sum of two terms with these coefficients: 6π‘₯βˆ’17π‘₯+12=6π‘₯βˆ’8π‘₯βˆ’9π‘₯+12.

Next, we separate the expression into two binomials and factor each separately:

We observe that there is a shared binomial factor of (3π‘₯βˆ’4) in the two halves of this expression, so factoring by this, we obtain 6π‘₯βˆ’8π‘₯βˆ’9π‘₯+12=(3π‘₯βˆ’4)(2π‘₯βˆ’3).

Hence, the fully factored form of the trinomial 6π‘₯βˆ’17π‘₯+12π‘₯ is π‘₯(3π‘₯βˆ’4)(2π‘₯βˆ’3).

The examples we have considered so far have concerned trinomials involving only a single variable. In our next example, we will consider how to factor a trinomial involving two variables.

Example 4: Factoring a Simple Two-Variable Trinomial Expression into a Product of Binomials

Factorize fully π‘₯βˆ’15π‘₯𝑦+54π‘¦οŠ¨οŠ¨.

Answer

This trinomial involves two variables: π‘₯ and 𝑦. To identify how we can factor this trinomial, let us consider its structure:

  • The first term is π‘₯, which is equivalent to π‘₯ multiplied by π‘₯.
  • The second term is a product of π‘₯ and 𝑦 (and a constant).
  • The third term is a constant multiplied by π‘¦οŠ¨, which is equivalent to 𝑦 multiplied by 𝑦.

This suggests that each binomial consists of a term in π‘₯ and a term in 𝑦. Hence, the factored form of the trinomial is (𝐴π‘₯+𝐡𝑦)(𝐢π‘₯+𝐷𝑦) for values of 𝐴, 𝐡, 𝐢, and 𝐷 to be determined.

In fact, as the coefficient of π‘₯ is 1, this suggests that both 𝐴 and 𝐢 are equal to 1 and, hence, the factored form of this trinomial is (π‘₯+𝐡𝑦)(π‘₯+𝐷𝑦).

We can confirm that this structure is correct by distributing the parentheses: (π‘₯+𝐡𝑦)(π‘₯+𝐷𝑦)=π‘₯(π‘₯+𝐷𝑦)+𝐡𝑦(π‘₯+𝐷𝑦)=π‘₯+𝐷π‘₯𝑦+𝐡π‘₯𝑦+𝐡𝐷𝑦=π‘₯+(𝐡+𝐷)π‘₯𝑦+𝐡𝐷𝑦.

This does have the same structure as the expression we are required to factor.

We can apply the same method as we did to factor monic quadratics to this problem. We begin by considering the factors of 54, which is the coefficient of π‘¦οŠ¨.

The factor pair we are looking for must also sum to the coefficient of π‘₯𝑦, which is βˆ’15. As the product of these factors should be positive 54, the two numbers we are looking for have the same sign. As their sum is negative, the two numbers must both be negative.

By considering our list of factor pairs, we identify that the numbers required are βˆ’6 and βˆ’9. Rewriting the middle term of the trinomial as the sum of two terms with these coefficients gives π‘₯βˆ’15π‘₯𝑦+54𝑦=π‘₯βˆ’6π‘₯π‘¦βˆ’9π‘₯𝑦+54𝑦.

Separating this expression into two binomials and factoring each gives

Finally, factoring by the shared binomial factor of (π‘₯βˆ’6𝑦) gives π‘₯βˆ’15π‘₯𝑦+54𝑦=(π‘₯βˆ’6𝑦)(π‘₯βˆ’9𝑦).

Let us now consider another example of factoring a trinomial involving two variables into a product of two binomials. This time, the trinomial will involve higher order powers. As in the previous example, we will apply the same method as we did for factoring quadratics once we have identified the structure of the factorized form.

Example 5: Factoring a Two-Variable Trinomial Expression into a Product of Binomials

Factorize fully 48π‘š+48π‘šπ‘›βˆ’15𝑛οŠͺ.

Answer

We begin by observing that the coefficients of all three terms are multiples of 3. Hence, we can factor the entire trinomial by 3 to give 48π‘š+48π‘šπ‘›βˆ’15𝑛=3ο€Ή16π‘š+16π‘šπ‘›βˆ’5𝑛.οŠͺοŠͺ

Now, consider the structure of this trinomial. The first term involves π‘šοŠͺ, which is equal to ο€Ήπ‘šο…οŠ¨οŠ¨, and the third term involves π‘›οŠ¨. The central term involves a product of π‘šοŠ¨ and 𝑛. This suggests that the factored form of this trinomial is ο€Ήπ΄π‘š+π΅π‘›ο…ο€ΉπΆπ‘š+π·π‘›ο…οŠ¨οŠ¨ for values of 𝐴, 𝐡, 𝐢, and 𝐷 to be determined.

We now need to find two numbers whose sum is equal to the coefficient of π‘šπ‘›(16) and whose product is equal to the product of the coefficients of the first and last terms (16Γ—βˆ’5=βˆ’80).

The factor pairs of 80 are as follows.

As the product should be βˆ’80, we require a factor pair with opposite signs such that their sum is equal to 16. The correct pair are 20 and βˆ’4.

Rewriting the second term in the trinomial as the sum of two terms with these coefficients gives 16π‘š+16π‘šπ‘›βˆ’5𝑛=16π‘š+20π‘šπ‘›βˆ’4π‘šπ‘›βˆ’5𝑛.οŠͺοŠͺ

Separating this four-term expression into two binomials and factoring each separately gives

Hence, the fully factored form of the trinomial is 48π‘š+48π‘šπ‘›βˆ’15𝑛=3ο€Ή4π‘šβˆ’π‘›ο…ο€Ή4π‘š+5𝑛.οŠͺ

The techniques we have encountered in this explainer can also be applied to problems in other areas of mathematics, such as geometry, or real-world problems. We now consider one final example in which we find the dimensions and perimeter of a rectangle given a trinomial algebraic expression for its area.

Example 6: Finding the Dimensions of a Rectangle by Factorization given Its Area as an Algebraic Expression

Using factorization, find, in terms of π‘₯, the dimensions of a rectangle given that its area is ο€Ή5π‘₯+12π‘₯+7ο…οŠ¨ cm2, and then find its perimeter when π‘₯=4.

Answer

The area of a rectangle is found by multiplying its length and width. We are given a quadratic expression for this area, and in order to find the rectangle’s dimensions, we are required to factor this expression into the product of two binomials.

The expression given is a nonmonic quadratic. As both the coefficient of π‘₯ and the constant term are prime numbers, one option is to determine the factored form of this quadratic using trial and error. The first term, 5π‘₯, factors as 5π‘₯ multiplied by π‘₯. The constant term, 7, factors as 7 multiplied by 1. As the coefficients of all three terms are positive, each term in the factored form must also be positive.

Using trial and error, we find that the factored form that produces the correct middle term of 12π‘₯ is (5π‘₯+7)(π‘₯+1).

Hence, the dimensions of the rectangle are (5π‘₯+7) cm and (π‘₯+1) cm.

If we did not wish to use trial and error, we could instead follow the more formal method of rewriting the trinomial as a four-term expression, by rewriting the middle term as the sum of two terms with coefficients whose product is equal to the product of the coefficient of π‘₯ and the constant term. Hence, we would require two numbers whose product is equal to 35(5Γ—7) and whose sum is equal to 12. These two numbers are 5 and 7, and hence our solution would be as follows: 5π‘₯+12π‘₯+7=5π‘₯+5π‘₯+7π‘₯+7=5π‘₯(π‘₯+1)+7(π‘₯+1)=(5π‘₯+7)(π‘₯+1), which is consistent with the factored form of the quadratic we have already found.

We are also required to find the perimeter of this rectangle when π‘₯=4. Evaluating the dimensions of the rectangle gives 5π‘₯+7=(5Γ—4)+7=27cm and (π‘₯+1)=4+1=5.cm

The perimeter of the rectangle, which is the sum of its four side lengths, is therefore equal to 27+27+5+5=64.cm

Let us finish by recapping some key points from this explainer.

Key Points

  • A monomial is a product of numbers and powers of variables.
  • A binomial expression is the sum or difference of two monomials.
  • A trinomial expression is the sum or difference of three monomials.
  • To factor a nonmonic quadratic of the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, where π‘Ž is not equal to 0, 1, or βˆ’1, we perform the following steps:
    • List the factor pairs of π‘Žπ‘.
    • If π‘Žπ‘ is positive, the two numbers will have the same sign, whereas if π‘Žπ‘ is negative, the two numbers will have opposite signs.
    • Look for a factor pair that, with the correct combination of signs, sum to the coefficient of π‘₯(𝑏).
    • Rewrite the middle term in the trinomial (𝑏π‘₯) as the sum of terms with coefficients equal to the factor pair found.
    • Separate the new four-term expression into two binomials and factor each.
    • Look for a shared binomial factor to factor the entire expression by.
  • Factoring monic quadratics of the form π‘₯+𝑏π‘₯+π‘οŠ¨ is a special case of the above, with π‘Ž equal to 1 (and hence π‘Žπ‘=𝑐).
  • To factor a two-variable trinomial, first consider its structure and identify the structure of the binomial factors. The coefficients can be found using the same method used for nonmonic quadratics.
  • Some trinomials can be factored into the product of more than two terms by first taking out a common factor.

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