Factoring Polynomials: Definition, Factored Form & Examples

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$3 x y^{2} - 27 x^{3} y^{3} + 15 x^{2} y - 9 x y^{4}$

Doesn't this look a little lengthy and somewhat confusing? However, do you see what each term has in common? Let me give you a hint: each term is divisible by 3, x and y. Taking these out of this expression, we will get a shorter and more straightforward expression. Though, more on that later!

In this article, we will be exploring factoring polynomials, their definition, how to solve polynomial equations and examples of applications of this.

Factoring

We will now begin with the definition of factoring.

Factoring is a process in which we write a number or an expression as a product of its factors.

This is an important topic when dealing with complex polynomials. Let us take the following example:

Factor the value of 20.

Solution

There are many ways to break down the product of 20.

Take 2 x 10 = 20, for instance.

However, 10 is not a prime and so we can further break down this product, as below

2 x 2 x 5 = 20.

Now, we can no longer simplify the components that make up this product. Thus, it has been completely factorised.

In algebra, factoring a (higher degree) polynomial means that we are rewriting a polynomial as a product of lower degree polynomials. Essentially, what we are doing here is the opposite of the FOIL method. Given the product of two binomials (a + b)(c + d), recall that the FOIL method is given by

Foil method, Aishah Amri - StudySmarter Originals

FOIL stands for First, Outer, Inner and Last.

This is the result of expanding the product of the two binomials. We aim to simplify a polynomial so that it is displayed as a product of its lowest degree polynomials rather than multiplying it out.

Before we get into this topic, let us recall the following definitions.

A monomial is a polynomial with only one term. For example, 2x.

A binomial is the sum of two monomials. For example, 3x + 1.

A trinomial is the sum of three monomials. For example, x + 2y + 5.

Why is Factoring Important?

In most cases, factoring plays an important role in simplifying an expression. This allows you to solve a particular equation in a more efficient way. Furthermore, factoring will help you understand the behaviour of a polynomial expression when graphing is required. We can also solve equations through factoring polynomials by identifying their solutions. The following are the learning objectives for this section:

Factoring a polynomial
Solve polynomial equations via factoring

Here, we shall be introduced to three types of factoring methods, namely:

Greatest common factor (GCF)
Quadratic trinomials
Grouping

Greatest Common Factor

The Greatest Common Factor (GCF) is the highest common monomial shared between all the components of a polynomial. When factoring polynomials, it is important to start by using this method to avoid dealing with larger numbers.

To perform this method, we first seek the GCF and factor it out of the polynomial. Fundamentally, we are carrying out the reverse of the distributive law, or in notation form:

$a b + b c = b (a + c)$

The general case for factoring polynomials using the GCF is:

$a^{3} b^{2} + 2 a^{2} b - 4 a b^{2} = a b (a^{2} b + 2 a - 4 b)$

Notice that all the terms on the left-hand side of the general form above have the common factor ab. Below are some worked examples to demonstrate this factoring method:

Factor out the GCF from the following expression,

$15 x^{4} - 20 x^{3} + 35 x^{2}$

Solution

Here, we can factor out 5 and x² from every term. In doing so, we obtain

$15 x^{4} - 20 x^{3} + 35 x^{2} = 5 x^{2} (3 x^{2} - 4 x + 7)$

Factor out the GCF from the following expression,

$x^{3} y^{2} + 3 x^{4} y + 5 x^{5} y^{3}$

Solution

Notice that each term contains both x³and y. Thus, factoring these out yields

$x^{3} y^{2} + 3 x^{4} y + 5 x^{5} y^{3} = x^{3} y (y + 3 x + 5 x^{2} y^{2})$

Factor out the GCF from the following expression,

$9 x^{2} (2 x + 7) - 12 x (2 x + 7)$

Solution

Firstly, observe that each term contains the binomial (2x + 7). Factoring this out yields

$9 x^{2} (2 x + 7) - 12 x (2 x + 7) = (2 x + 7) (9 x^{2} - 12 x)$

Now notice that we can further take out 3x from (9x² - 12x). This will give us the final factorized form as

$(2 x + 7) (9 x^{2} - 12 x) = 3 x (2 x + 7) (3 x - 4)$

Try verifying the factorized results using the FOIL method. Does it give the same expression?

Quadratic Trinomials

A second-degree polynomial is known as a quadratic. This means that the largest exponent in the polynomial is a 2. Typically, they take on the form of a quadratic trinomial as there are three terms, that is:

$a c x^{2} + (a d + b c) x + b d$

For this method of factoring, we seek to factor quadratic trinomials into a product of first-degree binomials. There are five steps to consider when applying this method:

Step 1: Write down a pair of parenthesis: ( ) ( )

Step 2: Deduce the product of the first term in the trinomial and add it into the parenthesis as: $(a x + □) (c x + □)$

Step 3: Look for factors of the third term of the trinomial. This should be in the form of a pair, such as $(b) (d) = b d$

Step 4: Use the trial and error method to identify which pair of factors satisfy the quadratic polynomial. In other words, find the pair that add up to the middle term of the trinomial. Remember to plug in all possible pairs and every possible order to determine the correct pair of values.

Step 5: Once found, add these values into the complete factorized form: $(a x + b) (c x + d)$

This form of factorization can be rather lengthy and complicated. Mastering this technique comes with lots of practice! Once you get the hang of the trial and error method, it will be very straightforward. We shall make use of the general formula below when factoring quadratic trinomials:

$a c x^{2} + (a d + b c) x + b d = (a x + b) (c x + d)$

Let us now look at some worked examples that exercise this factoring method.

Factorize the quadratic trinomial below,

$x^{2} + x - 12$

Solution

Steps 1 and 2: We start by looking at the first term, x². Factoring this may look like the expression below

$x^{2} + x - 12 = (x + □) (x + □)$

since multiplying x by x gives us x².

Step 3: To determine the values that go into the blank spaces, we must find a pair of numbers that multiply to get –12 and add to get 1. We do this by looking for factors of the last term, –12. Listing this down, we get the following pairs.

(-1)(12), (1)(-12), (-2)(6), (2)(-6), (-3)(4), (3)(-4)

Step 4: We find that the pair (-3)(4) satisfies the polynomial.

Step 5: Writing this in the final factored form, we obtain

$x^{2} + x - 12 = (x - 3) (x + 4)$

Important note: To determine the right pair of components in the factored form, be sure to plug each pair of factors into both possible forms and orders when using the trial and error method.

Factorize the quadratic trinomial below,

$6 x^{2} + 11 x - 10$

Solution

Steps 1 and 2: Looking at the first term, we have two possible factored forms,

$6 x^{2} + 11 x - 10 = (3 x + □) (2 x + □) a n d 6 x^{2} + 11 x - 10 = (6 x + □) (x + □)$

Step 3: To find the missing numbers, we observe the factors of the last term, –10.(–1)(10), (1)(–10), (–2)(5), (2)(–5)Step 4: From here, we need to look for a pair that also adds up to 11 on top of the criteria that its product equates to -10.

Step 5: With some trial and error, we obtain the final factorized form as

$6 x^{2} + 11 x - 10 = (3 x - 2) (2 x + 5)$

Factorize the quadratic trinomial below,

$3 x^{2} + 2 x - 8$

Solution

Steps 1 and 2: We can see that the first term takes on the factorized form

$3 x^{2} + 2 x - 8 = (3 x + □) (x + □)$

Step 3: We must find a pair of numbers that multiply to get –8 and add up to get 2. Looking at the final term, –8, we can list down its factors as

(-1)(8), (1)(-8), (-2)(4), (2)(-4)

Step 4: Plugging these into the blank spaces one by one tells us that (-4)(2) is the correct pair.

Step 5: In doing so, our final factorized form becomes

$3 x^{2} + 2 x - 8 = (3 x - 4) (x + 2)$

Grouping

The grouping method is often used when we encounter a polynomial with four terms. There are three steps to this method:

Step 1: Group the polynomial into two sets of two terms. This is usually done by splitting the first two terms and the last two terms of the polynomial as, $(a x + b x) + (a y + b y)$

Step 2: Factorize each group using the GCF factoring method: $x (a + b) + y (a + b)$

Step 3: If we find a common polynomial between the two factorized groups, apply the GCF factoring method again to factor it out: $(a + b) (x + y)$

The general case for factoring polynomials via grouping is as follows:

$a x + b x + a y + b y = x (a + b) + y (a + b) = (a + b) (x + y)$

We shall now observe the following worked examples that employ this factoring method.

Use the grouping method to factorize the polynomial $x^{3} + 2 x^{2} - 3 x - 6$

Solution

Step 1: We start by grouping the first two terms and last two terms as below.

$(x^{3} + 2 x^{2}) + (- 3 x - 6)$

Step 2: Notice that we can factor out x² from the first group and –3 from the second group.

$x^{2} (x + 2) + (- 3) (x + 2)$

Step 3: Now observing that both groups have a common binomial (x + 2), we can factor this out to obtain

$x^{3} + 2 x^{2} - 3 x - 6 = (x + 2) (x^{2} - 3)$

Use the grouping method to factorize the polynomial $3 x y - 21 y + 5 x - 35$

Solution

Step 1: As before, we start by grouping the first two terms and last two terms as below.

$(3 x y - 21 y) + (5 x - 35)$

Step 2: Notice that we can factor out 3y from the first group and 5 from the second group.

$3 y (x - 7) + 5 (x - 7)$

Step 3: Now observing that both groups have a common binomial (x – 7), we can factor this out to obtain

$3 x y - 21 y + 5 x - 35 = (x - 7) (3 y + 5)$

Use the grouping method to factorize the polynomial $x^{5} + x - 2 x^{4} - 2$

Solution

Step 1: Start by grouping the first two terms and last two terms as below.

$(x^{5} + x) + (- 2 x^{4} - 2)$

Step 2: Notice that we can factor out x from the first group and –2 from the second group.

$x (x^{4} + 1) + (- 2) (x^{4} + 1)$

Step 3: Now observing that both groups have a common binomial (x⁴ + 1), we can factor this out to obtain

$x^{5} + x - 2 x^{4} - 2 = (x^{4} + 1) (x - 2)$

Grouping 3-Term Polynomials

Grouping is an amazing way to factor 3 term polynomials as well! Say we are given the expression below

$a + b + c$

Here, we shall use the trial and error method to find the 2 numbers that add to b and multiply to get ac. This makes a 4 term polynomial as below.

$a + \underset{= b}{\underset{⏟}{a B + c B}} + c$

Finally, we shall apply the grouping method to simplify the expression as before

$a (1 + B) + c (1 + B) = (a + c) (1 + B)$

Let us look at the example below.

Use the grouping method to solve the trinomial, $x^{2} - 8 x - 9$

Solution

Here, a = 1, b = –8 and c = –9. The product of ac is –9. We need a pair of numbers that add up to –8 and yield –9 when multiplied together. Our 4-term polynomial should look like this:

$x^{2} + □ x + □ x - 9$

To find the missing numbers, we shall look at the middle term, –8. Notice that if we add 1 to –9 we obtain –8. The product of 1 and –9 is –9, that is ac. We thus have the following 4-term polynomial.

$x^{2} + x - 9 x - 9$

Now, let us group the first two terms and the last two terms as below.

$(x^{2} + x) + (- 9 x - 9)$

Notice that we can factor out x from the first group and –9 from the second group.

$x (x + 1) + (- 9) (x + 1)$

Now observing that both groups have a common binomial (x + 1), we can factor this out to obtain

$x^{2} - 8 x - 9 = (x + 1) (x - 9)$

Factoring Polynomials of Higher Degree

So far, the polynomials that we have dealt with have a degree of two. What if we need to factorize polynomials of degree higher than that?

When tackling such expressions, there is no particular method in factorizing them. However, we can indeed apply the techniques introduced in this lesson. It is helpful to become well acquainted with these factoring methods so that you can easily seek common patterns when solving them. Let us look at some worked examples to show this.

Completely factorize the expression $3 x^{5} - 7 x^{4} - 26 x^{3}$

Solution

First, notice that we can factor out x³ from each term.

$x^{3} (3 x^{2} - 7 x - 26)$

Now we have the quadratic trinomial $3 x^{2} - 7 x - 26$ . From the first term the factorized form will take on the structure.

$3 x^{5} - 7 x^{4} - 26 x^{3} = x^{3} (3 x + □) (x + □)$

We need to find a pair of numbers that multiply to get –26 and add to get –7. Let us look at the factors of the term –26.

(-1)(26), (1)(-26), (-2)(13), (2)(-13)

Using trial and error, we find that the pair (–13)(2) satisfies the criteria for this factored form, and so the fully factorised form for this polynomial is

$3 x^{5} - 7 x^{4} - 26 x^{3} = x^{3} (3 x - 13) (x + 2)$

Fully factorize the expression $x^{3} - 2 x^{2} - 9 x + 18$

Solution

We begin by using the grouping method to factorize this. Grouping the first two terms and last two terms, we obtain

$(x^{3} - 2 x^{2}) + (- 9 x + 18)$

We can factor out x² from the first group and –9 from the second group.

$x^{2} (x - 2) + (- 9) (x - 2)$

We now have the common factor of (x – 2) in both groups. Factoring this out yields,

$x^{3} - 2 x^{2} - 9 x + 18 = (x - 2) (x^{2} - 9)$

Now, you may think that we have completed factoring our expression here. However, this is not the case. Observe the binomial $(x^{2} - 9)$ . This is an example of a perfect square binomial. We will look at this in more detail on the topic of Special Products.

To avoid confusion, we shall treat this as a quadratic trinomial where the middle term is essentially zero. By the first term, we can see that the factorized form must look like the following

$x^{3} - 2 x^{2} - 9 x + 18 = (x - 2) (x + □) (x + □)$

Now we need to look for a pair of numbers that multiply to get –9 and add up to get 0. The pair (–3)(3) is the most possible answer here. Thus, we obtain

$x^{3} - 2 x^{2} - 9 x + 18 = (x - 2) (x - 3) (x + 3)$

Factorize the expression $x^{4} - 13 x^{2} + 36$

Solution

First, notice that this expression looks similar to a quadratic trinomial. Let us use grouping to solve this. This 3-term polynomial will take the form of the 4-term polynomial below

$x^{4} + □ x^{2} + □ x^{2} + 36$

We must now find 2 numbers that add up to get –13 and multiply to get 36. By trial and error, we find that the pair –4 and –9 satisfy this criterion.

$x^{4} - 4 x^{2} - 9 x^{2} + 36$

Grouping the first two terms and last two terms, we obtain

$(x^{4} - 4 x^{2}) + (- 9 x^{2} + 36)$

Factoring x² from the first group and –9 from the second group, we get

$x^{2} (x^{2} - 4) + (- 9) (x^{2} - 4)$

Notice that we can factor out (x² - 4) from both groups, yielding

$x^{4} - 13 x^{2} + 36 = (x^{2} - 4) (x^{2} - 9)$

But we are not done yet! Just like the previous examples, we have the perfect square binomials $(x^{2} - 4) a n d (x^{2} - 9)$ . Thus, we can further factorize this. Using the same principle as before, we finally obtain

$x^{4} - 13 x^{2} + 36 = (x - 2) (x + 2) (x - 3) (x + 3)$

Solving Polynomial Equations

Now that we have mastered factoring polynomials, we can move on to solving them! Say we have the standard form of a polynomial equated to zero as below:

$a c x^{2} + (a d + b c) x + b d = 0$

As before, we know that the factorized form for this polynomial is:

$(a x + b) (c x + d) = 0$

Notice that the product of these two factors is zero if at least one of the factors is zero, that is either $(a x + b) = 0 o r (c x + d) = 0$ . This is known as the Zero Product Property, stated below:

The Zero Product Property

If ab = 0 then a = 0 or b = 0 (or both a = 0, b = 0)

Applying this property, we can solve the general form of a factorized polynomial as above, by identifying that:

$a x + b = 0 a n d c x + d = 0$

Rearranging this in terms of x, we have two solutions:

$x = - \frac{b}{a} a n d x = - \frac{d}{c}$

Do you see how factoring plays an important role in solving equations? Solving the equation via factoring provides the x-intercepts of the graph for a given equation. Let us return to the last three worked examples we did to show this.

Find the solutions to the polynomial below using the Zero Product Property

$3 x^{5} - 7 x^{4} - 26 x^{3} = 0$

Solution

From our solution above, the factorized form is

$x^{3} (3 x - 13) (x + 2) = 0$

As we have a product of 3 factors, we must have 3 solutions. Applying the Zero Product Property, we obtain

$x^{3} = 0 3 x - 13 = 0 x + 2 = 0$

Solving these for x, we obtain 3 real roots

$x = - 2, x = 0 a n d x = \frac{13}{3}$

Find the solutions to the polynomial below using the Zero Product Property

$x^{3} - 2 x^{2} - 9 x + 18 = 0$

Solution

The factorized form is

$(x - 2) (x - 3) (x + 3) = 0$

As we have a product of 3 factors, we must have 3 solutions. Applying the Zero Product Property, we obtain

$x - 2 = 0 x - 3 = 0 x + 3 = 0$

Solving these for x, we obtain 3 real roots

$x = - 3, x = 2 a n d x = 3$

The graph is shown below.

Example 1, Aishah Amri - StudySmarter Originals

Find the solutions to the polynomial below using the Zero Product Property

$x^{4} - 13 x^{2} + 36 = 0$

Solution

We have the factorized form as

$(x - 2) (x + 2) (x - 3) (x + 3) = 0$

As we have a product of 4 factors, we must have 4 solutions. Applying the Zero Product Property, we obtain

$x - 2 = 0 x + 2 = 0 x - 3 = 0 x + 3 = 0$

Solving these for x, we obtain 4 real roots

$x = - 3, x = - 2, x = 2 a n d x = 3$

The graph is shown below.

Example 2(1), Aishah Amri - StudySmarter Originals

Zooming into the x-intercepts, we have

Example 2(2), Aishah Amri - StudySmarter Originals

It is helpful to note that when solving repeated factor equations, we only obtain one solution. For example,

Solve ${(x - 7)}^{2} = 0$

Solution

Square root both sides.

$\sqrt{{(x - 7)}^{2}} = \sqrt{0} \Rightarrow x - 7 = 0$

The square root of zero is zero. Solving this yields

$x = 7$

Thus, we only have one real solution, as stated. The graph is shown below.

Example 3, Aishah Amri - StudySmarter Originals

Factoring Polynomials - Key takeaways

Factoring a polynomial means is a process of rewriting a polynomial as a product of lower degree polynomials.
Factoring plays an important role in simplifying an expression.
The Zero Product Property states that if ab = 0 then a = 0 or b = 0 (or both a = 0, b = 0).
Use the Zero Product Property to solve a factored polynomial.
The table below summarizes the factoring techniques throughout this lesson. Remember to familiarize yourself with them!

Type of Factoring	Technique
Factoring out the GCF	$a^{3} b^{2} + 2 a^{2} b - 4 a b^{2} = a b (a^{2} b + 2 a - 4 b)$
Quadratic trinomials	$a c x^{2} + (a d + b c) x + b d = (a x + b) (c x + d)$
Factoring by grouping	$a x + b x + a y + b y = x (a + b) + y (a + b) = (a + b) (x + y)$

Flashcards in Factoring Polynomials 1

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When factoring polynomials, what is the first method you should consider to avoid dealing with larger numbers?

Factoring out the greatest common factor.

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Frequently Asked Questions about Factoring Polynomials

How to factor polynomials?

we factor polynomials by breaking down the expression into a product of linear equations

How to factor polynomials with 4 terms?

we can factor polynomials with 4 terms using the grouping method

What is the factored form of the polynomial?

a factored form of a polynomial is an expression that is written as a product of linear terms with its factors

What are ways to factor polynomials?

we can factor polynomials by the greatest common factor, the grouping method and identifying quadratic trinomials

What are the linear factors of a polynomial?

linear factors of a polynomial are first-degree polynomials that make up the product of a higher degree polynomial

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Factoring Polynomials

StudySmarter Editorial Team

Factoring

Why is Factoring Important?

Greatest Common Factor

Quadratic Trinomials

Grouping

Grouping 3-Term Polynomials

Factoring Polynomials of Higher Degree

Solving Polynomial Equations

The Zero Product Property

Factoring Polynomials - Key takeaways

Flashcards in Factoring Polynomials 1

Learn faster with the 1 flashcards about Factoring Polynomials

Frequently Asked Questions about Factoring Polynomials

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Factoring Polynomials

StudySmarter Editorial Team

Factoring

Why is Factoring Important?

Greatest Common Factor

Quadratic Trinomials

Grouping

Grouping 3-Term Polynomials

Factoring Polynomials of Higher Degree

Solving Polynomial Equations

The Zero Product Property

Factoring Polynomials - Key takeaways

Flashcards in Factoring Polynomials 1

Learn faster with the 1 flashcards about Factoring Polynomials

Frequently Asked Questions about Factoring Polynomials

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