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Lesson Explainer: Factoring Algebraic Expressions Mathematics

In this explainer, we will learn how to write algebraic expressions as a product of irreducible factors.

Factoring an algebraic expression is the reverse process of expanding a product of algebraic factors. We can use the process of expanding, in reverse, to factor many algebraic expressions.

For example, we can expand π‘₯(π‘₯+2) by distributing the factor of π‘₯:

If we write this equation in reverse, then we have π‘₯+2π‘₯=ο€Ήπ‘₯Γ—π‘₯+ο€Ήπ‘₯Γ—2=π‘₯ο€Ήπ‘₯+2.

Starting with π‘₯+2π‘₯, we note that both terms share a factor of π‘₯. Then, we take this shared factor out to get π‘₯+2π‘₯=π‘₯(π‘₯+2).

We can follow this same process to factor any algebraic expression in which every term shares a common factor.

For example, let’s factor the expression 4π‘₯𝑦+2π‘₯π‘¦βˆ’6π‘₯. We want to check for common factors of all three terms, which we can start doing by checking for common constant factors shared between the terms. We see that 4, 2, and 6 all share a common factor of 2. In fact, this is the greatest common factor of the three numbers.

Next, we can also check for common factors of powers of π‘₯. We see that all three terms have factors of π‘₯: 4π‘₯𝑦+2π‘₯π‘¦βˆ’6π‘₯. We cannot take out a factor of a higher power of π‘₯ since π‘₯ is the largest power in the three terms.

Finally, we can check for a common factor of a power of 𝑦. We note that the final term, βˆ’6π‘₯, has no factors of 𝑦, so we cannot take a factor of any power of 𝑦 out of the expression.

Therefore, we find that the common factors are 2 and π‘₯, which we can multiply to get 2π‘₯; this is the greatest common factor of the three terms.

We want to take the factor of 2π‘₯ out of the expression. Doing this separately for each term, we obtain 4π‘₯𝑦=2π‘₯Γ—2𝑦,2π‘₯𝑦=2π‘₯Γ—π‘₯𝑦,βˆ’6π‘₯=2π‘₯Γ—(βˆ’3).

Hence, we can factor the expression to get 4π‘₯𝑦+2π‘₯π‘¦βˆ’6π‘₯=2π‘₯ο€Ί2𝑦+π‘₯π‘¦βˆ’3ο†οŠ¨οŠ¨οŠ¨

We call 2π‘₯ the greatest common factor of the terms since we cannot take out any further factors.

In our first example, we will follow this process to factor an algebraic expression by identifying the greatest common factor of its terms.

Example 1: Factoring an Expression by Identifying the Greatest Common Factor

Factorize 12π‘₯π‘¦βˆ’8π‘₯π‘¦οŠ©οŠ¨οŠ© fully.

Answer

To factor the expression, we need to find the greatest common factor of the two terms. We can do this by finding the greatest common factor of the coefficients and each variable separately.

Let’s start with the coefficients. We want to find the greatest factor of 12 and 8. We can see that 12=4Γ—3 and 8=4Γ—2 and that 2 and 3 share no common factors other than 1. Thus, 4 is the greatest common factor of the coefficients.

We can now check each term for factors of powers of π‘₯. We see that the first term has a factor of π‘₯ and the second term has a factor of π‘₯: 12π‘₯π‘¦βˆ’8π‘₯𝑦.

We cannot take out more than the lowest power as a factor, so the greatest shared factor of a power of π‘₯ is just π‘₯.

Similarly, if we consider the powers of 𝑦 in each term, 12π‘₯π‘¦βˆ’8π‘₯𝑦, we see that every term has a power of 𝑦 and that the lowest power of 𝑦 is π‘¦οŠ¨. Therefore, the greatest shared factor of a power of 𝑦 is π‘¦οŠ¨.

We can multiply these together to find that the greatest common factor of the terms is 4π‘₯π‘¦οŠ¨.

Let’s factor 4π‘₯π‘¦οŠ¨ from each term separately.

For the first term, we have 12π‘₯𝑦=4π‘₯𝑦×3π‘₯.

For the second term, we have βˆ’8π‘₯𝑦=4π‘₯𝑦×(βˆ’2𝑦).

This allows us to take out the factor of 4π‘₯π‘¦οŠ¨ as follows: 12π‘₯π‘¦βˆ’8π‘₯𝑦=ο€Ή4π‘₯𝑦×3π‘₯+ο€Ή4π‘₯𝑦×(βˆ’2𝑦)=4π‘₯𝑦3π‘₯βˆ’2𝑦.

In our next example, we will factor an algebraic expression with three terms.

Example 2: Factoring an Expression with Three Terms

Factor the expression 6𝑝+3π‘βˆ’6π‘π‘žοŠ¨ completely.

Answer

To factor the expression, we need to find the greatest common factor of all three terms. We can do this by finding the greatest common factor of the coefficients and each variable separately.

Let’s start with the coefficients. We note that all three terms are divisible by 3 and no greater factor exists, so it is the greatest common factor of the coefficients.

We can now look for common factors of the powers of the variables. If we highlight the instances of the variable 𝑝, 6𝑝+3π‘βˆ’6π‘π‘ž, we see that all three terms share factors of 𝑝. The lowest power of 𝑝 is just 𝑝, so this is the greatest common factor of 𝑝 in the three terms.

If we highlight the factors of π‘ž, 6𝑝+3π‘βˆ’6π‘π‘ž, we see that there are terms with no factor of π‘ž. This means we cannot take out any factors of π‘ž.

Thus, the greatest common factor of the three terms is 3𝑝. Taking a factor of 3𝑝 out of the first term yields 6𝑝=3𝑝×2𝑝.

Taking a factor of 3𝑝 out of the second term gives us 3𝑝=3𝑝×1.

Taking a factor of 3𝑝 out of the third term produces βˆ’6π‘π‘ž=3𝑝×(βˆ’2π‘ž).

This allows us to take out the factor of 3𝑝 as follows: 6𝑝+3π‘βˆ’6π‘π‘ž=(3𝑝×2𝑝)+(3𝑝×1)+(3𝑝×(βˆ’2π‘ž))=3𝑝(2𝑝+1βˆ’2π‘ž).

We usually write the constants at the end of the expression, so we have 6𝑝+3π‘βˆ’6π‘π‘ž=3𝑝(2π‘βˆ’2π‘ž+1).

We can also examine the process of expanding two linear factors to help us understand the reverse process, factoring quadratic expressions. For example, if we expand (2π‘₯+3)(π‘₯+2), we get (2π‘₯+3)(π‘₯+2)=π‘₯(2π‘₯+3)+2(2π‘₯+3)=2π‘₯+3π‘₯+4π‘₯+6=2π‘₯+7π‘₯+6.

To reverse this process, we would start with 2π‘₯+7π‘₯+6 and work backward to write it as two linear factors. But how would we know to separate 7π‘₯ into 3π‘₯+4π‘₯? We can do this by noticing special qualities of 3 and 4, which are the coefficients of 3π‘₯ and 4π‘₯:

That is, we can see that the product of 3 and 4 is equal to the product of 2 and 6 (i.e., the π‘₯-coefficient and the constant coefficient) and that the sum of 3 and 4 is 7 (i.e., the π‘₯-coefficient). To put this in general terms, for a quadratic expression of the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, we have identified a pair of numbers π‘š and 𝑛 such that π‘Žπ‘=π‘šπ‘› and 𝑏=π‘š+𝑛.

By identifying pairs of numbers as shown above, we can factor any general quadratic expression. To see this, let’s consider the expansion of (𝑝π‘₯+π‘ž)(π‘Ÿπ‘₯+𝑠): (𝑝π‘₯+π‘ž)(π‘Ÿπ‘₯+𝑠)=π‘π‘Ÿπ‘₯+π‘žπ‘Ÿπ‘₯+𝑝𝑠π‘₯+π‘žπ‘ =π‘π‘Ÿπ‘₯+(π‘žπ‘Ÿ+𝑝𝑠)π‘₯+π‘žπ‘ .

Let’s compare this result to the general form of a quadratic expression π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨. We can see that π‘Ž=π‘π‘Ÿ, 𝑏=π‘žπ‘Ÿ+𝑝𝑠, and 𝑐=π‘žπ‘ , so we have π‘Žπ‘=π‘π‘Ÿπ‘žπ‘ =(π‘žπ‘Ÿ)(𝑝𝑠),𝑏=π‘žπ‘Ÿ+𝑝𝑠.

In other words, π‘š=π‘žπ‘Ÿ and 𝑛=𝑝𝑠, which are the coefficients of the π‘₯-terms that appear in the expansion; they are two numbers that multiply to make π‘Žπ‘ and sum to give 𝑏.

With this property in mind, let’s examine a general method that will allow us to factor any quadratic expression.

How To: Factoring a Single-Variable Quadratic Polynomial

We can factor a quadratic polynomial of the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨ using the following steps:

  1. Calculate π‘Žπ‘ and list its factor pairs; find the pairs of numbers π‘š and 𝑛 such that π‘šπ‘›=π‘Žπ‘.
  2. Add the factors of π‘Žπ‘ together to find two factors that add to give 𝑏. Let’s say π‘š+𝑛=𝑏.
  3. Rewrite the π‘₯-term using these factors. So, we have π‘Žπ‘₯+π‘šπ‘₯+𝑛π‘₯+π‘οŠ¨.
  4. Factor the first two terms and final two terms separately.
  5. Take out the common factor.

Let’s see this method applied to an example. First of all, we will consider factoring a monic quadratic expression (one where the π‘₯-coefficient is 1).

Example 3: Factoring a Quadratic Expression

Factor π‘₯+8π‘₯+12.

Answer

We are asked to factor a quadratic expression with leading coefficient 1. We can do this by finding two numbers whose sum is the coefficient of π‘₯, 8, and whose product is the constant, 12.

One way of finding a pair of numbers like this is to list the factor pairs of 12:

We see that 2+6=8 and 2Γ—6=12. We use this to rewrite the π‘₯-term in the quadratic: π‘₯+8π‘₯+12=π‘₯+6π‘₯+2π‘₯+12.

We now note that the first two terms share a factor of π‘₯ and the final two terms share a factor of 2. Taking out these factors yields π‘₯+6π‘₯+2π‘₯+12=π‘₯(π‘₯+6)+2(π‘₯+6).

We can now note that both terms share a factor of π‘₯+6. Taking out this factor gives π‘₯(π‘₯+6)+2(π‘₯+6)=(π‘₯+6)(π‘₯+2).

There are many other methods we can use to factor quadratics. For example, we can expand a product of the form (π‘₯+𝑦)(π‘₯βˆ’π‘¦) to obtain (π‘₯+𝑦)(π‘₯βˆ’π‘¦)=π‘₯βˆ’π‘₯𝑦+π‘₯π‘¦βˆ’π‘¦=π‘₯βˆ’π‘¦.

We note that the terms π‘₯𝑦 and βˆ’π‘₯𝑦 sum to give zero in the expasion, which leads to an expression with only two terms. We call this resulting expression a difference of two squares, and by applying the above steps in reverse, we arrive at a way to factor any such expression.

Property: Factoring a Difference of Two Squares

An expression of the form π‘₯βˆ’π‘¦οŠ¨οŠ¨ is called a difference of two squares. We can factor this as π‘₯βˆ’π‘¦=(π‘₯+𝑦)(π‘₯βˆ’π‘¦).

In our next example, we will use this property of a factoring a difference of two squares to factor a given quadratic expression.

Example 4: Factoring the Difference of Two Squares

Factor the expression π‘₯βˆ’49.

Answer

We first note that the expression we are asked to factor is the difference of two squares since 49=7. We can recall that π‘₯βˆ’π‘¦=(π‘₯+𝑦)(π‘₯βˆ’π‘¦).

Therefore, taking 𝑦=7, we have π‘₯βˆ’49=π‘₯βˆ’7=(π‘₯+7)(π‘₯βˆ’7).

In our next example, we will see how to apply this process to factor a polynomial using a substitution.

Example 5: Factoring a Polynomial Using a Substitution

Factorize fully π‘¦βˆ’5π‘¦βˆ’14οŠͺ.

Answer

We want to fully factor the given expression; however, we can see that the three terms share no common factor and that this is not a quadratic expression since the highest power of 𝑦 is 4.

We can rewrite the given expression as a quadratic using the substitution π‘₯=π‘¦οŠ¨. To see this, we rewrite the expression using the laws of exponents: π‘¦βˆ’5π‘¦βˆ’14=ο€Ήπ‘¦ο…βˆ’5π‘¦βˆ’14.οŠͺ

Using the substitution π‘₯=π‘¦οŠ¨ gives us ο€Ήπ‘¦ο…βˆ’5π‘¦βˆ’14=π‘₯βˆ’5π‘₯βˆ’14.

We can now factor the quadratic by noting it is monic, so we need two numbers whose product is βˆ’14 and whose sum is βˆ’5. We can find these by considering the factors of βˆ’14: βˆ’14,1βˆ’7,2βˆ’2,7βˆ’1,14.

We see that βˆ’7Γ—2=βˆ’14 and βˆ’7+2=βˆ’5, so we will use these values to split the π‘₯-term: π‘₯βˆ’5π‘₯βˆ’14=π‘₯βˆ’7π‘₯+2π‘₯βˆ’14.

We take out the shared factor of π‘₯ in the first two terms and the shared factor of 2 in the final two terms to obtain π‘₯βˆ’7π‘₯+2π‘₯βˆ’14=π‘₯(π‘₯βˆ’7)+2(π‘₯βˆ’7).

Now, we can take out the shared factor of π‘₯βˆ’7 from the two terms to get π‘₯(π‘₯βˆ’7)+2(π‘₯βˆ’7)=(π‘₯+2)(π‘₯βˆ’7).

We could leave our answer like this; however, the original expression we were given was in terms of 𝑦. So, we will substitute π‘₯=π‘¦οŠ¨ into the factored expression to get π‘¦βˆ’5π‘¦βˆ’14=𝑦+2ο…ο€Ήπ‘¦βˆ’7.οŠͺ

In our next example, we will fully factor a nonmonic quadratic expression.

Example 6: Factoring a Nonmonic Quadratic Expression

Factorize fully 6π‘₯βˆ’19π‘₯+10.

Answer

We first recall that to factor a quadratic in the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, we need to find two numbers whose product is π‘Žπ‘ and whose sum is 𝑏. In our case, we have π‘Ž=6, 𝑏=βˆ’19, and 𝑐=10, so we want two numbers that sum to give βˆ’19 and multiply to give 6Γ—10=60. Since the numbers sum to give βˆ’19, one of the numbers must be negative, so we will only check the factor pairs of 60 that contain negative factors:

We see that βˆ’15 and βˆ’4 are the factors of 60 whose sum is βˆ’19.

We use these two numbers to rewrite the π‘₯-term as follows: 6π‘₯βˆ’19π‘₯+10=6π‘₯βˆ’15π‘₯βˆ’4π‘₯+10.

Then, we can take out the shared factor of 3π‘₯ in the first two terms and the shared factor of 2 in the final two terms to get 6π‘₯βˆ’15π‘₯βˆ’4π‘₯+10=3π‘₯(2π‘₯βˆ’5)+2(βˆ’2π‘₯+5).

To make the two terms share a factor, we need to take a factor of βˆ’1 out of the second term to obtain 3π‘₯(2π‘₯βˆ’5)+2(βˆ’2π‘₯+5)=3π‘₯(2π‘₯βˆ’5)βˆ’2(2π‘₯βˆ’5).

Finally, we take out the shared factor of 2π‘₯βˆ’5: 3π‘₯(2π‘₯βˆ’5)βˆ’2(2π‘₯βˆ’5)=(2π‘₯βˆ’5)(3π‘₯βˆ’2).

In our final example, we will apply this process to fully factor a nonmonic cubic expression.

Example 7: Factoring a Nonmonic Cubic Expression

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

Answer

We note that this expression is cubic since the highest nonzero power of π‘₯ is π‘₯. When factoring cubics, we should first try to identify whether there is a common factor of π‘₯ we can take out. Since all three terms share a factor of π‘₯, we can take out this factor to yield 6π‘₯βˆ’17π‘₯+12π‘₯=π‘₯ο€Ή6π‘₯βˆ’17π‘₯+12.

We can factor the quadratic further by recalling that to factor π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, we need to find two numbers whose product is π‘Žπ‘ and whose sum is 𝑏. In our case, we have π‘Ž=6, 𝑏=βˆ’17, and 𝑐=12, so we want two numbers that sum to give βˆ’17 and multiply to give 6Γ—12=72. Since the numbers sum to give βˆ’17, one of the numbers must be negative, so we will only check the factor pairs of 72 that contain negative factors:

We find that these numbers are βˆ’8 and βˆ’9.

We use these two numbers to rewrite the π‘₯-term as follows: 6π‘₯βˆ’17π‘₯+12=6π‘₯βˆ’9π‘₯βˆ’8π‘₯+12.

Then, we can take out the shared factor of 3π‘₯ in the first two terms and the shared factor of 4 in the final two terms to get 6π‘₯βˆ’9π‘₯βˆ’8π‘₯+10=3π‘₯(2π‘₯βˆ’3)+4(βˆ’2π‘₯+3).

To make the two terms share a factor, we need to take a factor of βˆ’1 out of the second term to obtain 3π‘₯(2π‘₯βˆ’3)+4(βˆ’2π‘₯+3)=3π‘₯(2π‘₯βˆ’3)βˆ’4(2π‘₯βˆ’3).

Finally, we take out the shared factor of 2π‘₯βˆ’3: 3π‘₯(2π‘₯βˆ’3)βˆ’4(2π‘₯βˆ’3)=(3π‘₯βˆ’4)(2π‘₯βˆ’3).

Hence, 6π‘₯βˆ’17π‘₯+12π‘₯=π‘₯(3π‘₯βˆ’4)(2π‘₯βˆ’3).

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The greatest common factor of an algebraic expression is the greatest common factor of the coefficients multiplied by each variable raised to the lowest exponent in which it appears in any term.
  • We can factor an algebraic expression by checking for the greatest common factor of all of its terms and taking this factor out.
  • An expression of the form π‘₯βˆ’π‘¦οŠ¨οŠ¨ is called a difference of two squares. We can factor this as π‘₯βˆ’π‘¦=(π‘₯+𝑦)(π‘₯βˆ’π‘¦).
  • We can factor a quadratic in the form π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨ by finding two numbers whose product is π‘Žπ‘ and whose sum is 𝑏. We use these two numbers to rewrite the π‘₯-term and then factor the first pair and final pair of terms. Finally, we factor the whole expression.
  • If we are asked to factor a cubic or higher-degree polynomial, we should first check if each term shares any common factors of the variable to simplify the expression.

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