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# Factoring Polynomials

When it comes to solving polynomials, it is always handy to express them in a more concise way. That way, we can spot patterns that allow us to solve such polynomials easily. What if I told you that there is a nifty trick you can use to do just that? That's right! This nifty trick is called factoring. Let me give you a scenario. Say we are given the expression below,

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:

1. Greatest common factor (GCF)

3. 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:

The general case for factoring polynomials using the GCF is:

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,

Solution

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

Factor out the GCF from the following expression,

Solution

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

Factor out the GCF from the following expression,

Solution

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

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

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

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:

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:

Step 3: Look for factors of the third term of the trinomial. This should be in the form of a pair, such as

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:

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:

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

Solution

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

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

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.

Solution

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

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

Solution

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

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

## 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,

Step 2: Factorize each group using the GCF factoring method:

Step 3: If we find a common polynomial between the two factorized groups, apply the GCF factoring method again to factor it out:

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

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

Use the grouping method to factorize the polynomial

Solution

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

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

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

Use the grouping method to factorize the polynomial

Solution

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

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

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

Use the grouping method to factorize the polynomial

Solution

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

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

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

### Grouping 3-Term Polynomials

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

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.

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

Let us look at the example below.

Use the grouping method to solve the trinomial,

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:

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.

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

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

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

## 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

Solution

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

Now we have the quadratic trinomial . From the first term the factorized form will take on the structure.

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

Fully factorize the expression

Solution

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

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

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

Now, you may think that we have completed factoring our expression here. However, this is not the case. Observe the binomial . 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

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

Factorize the expression

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

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.

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

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

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

But we are not done yet! Just like the previous examples, we have the perfect square binomials . Thus, we can further factorize this. Using the same principle as before, we finally obtain

### 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:

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

Notice that the product of these two factors is zero if at least one of the factors is zero, that is either . 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:

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

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

Solution

From our solution above, the factorized form is

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

Solving these for x, we obtain 3 real roots

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

Solution

The factorized form is

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

Solving these for x, we obtain 3 real roots

The graph is shown below.

Example 1, Aishah Amri - StudySmarter Originals

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

Solution

We have the factorized form as

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

Solving these for x, we obtain 4 real roots

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

Solution

Square root both sides.

The square root of zero is zero. Solving this yields

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 Quadratic trinomials Factoring by grouping

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

we can factor polynomials with 4 terms using the grouping method

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

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

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

## Final Factoring Polynomials Quiz

Question

How many solutions do we obtain when we factorise a perfect square trinomial? What are the properties of these solutions?

We get one solution. This is a repeated real root.

Show question

Question

How many solutions do we obtain from either the sum or difference of two cubes? What are the properties of these solutions?

We get three solutions. This includes one real root and two distinct complex roots.

Show question

Question

What does it mean to factor a number or expression?

Factoring is the process of writing a number or expression as a product of primes.

Show question

Question

What does it mean to factor a polynomial?

Factoring a polynomial means that we are rewriting a polynomial as a product of lower degree polynomials.

Show question

Question

List three reasons why factoring is important?

1. Simplifying an expression.
2. Finding roots of a polynomial.
3. Identifying the behaviour of a polynomial when graphing is required.

Show question

Question

What is the Greatest Common Factor of a polynomial?

The GCF is the highest common monomial shared between all the components of a polynomial.

Show question

Question

What is the general case for factoring polynomials using the GCF method?

$$a^3b^2 + 2a^2b - 4ab^2 = ab(a^2b+2a-4b)$$.

Show question

Question

What is the general formula when factoring quadratic trinomials?

$$acx^2 + (ad+bc)x + bd = (ax+b)(cx+d)$$.

Show question

Question

Can you list down the five steps for factoring quadratic trinomials?

2. Deduce the product of the first term in the trinomial.
3. Look for factors of the third term of the trinomial.
4. Use trial and error to identify which pair of factors add up to satisfy the middle term.
5. Add these values into the complete factorised form.

Show question

Question

What is the general formula of the grouping method for factoring polynomials?

$$ax+bx+ay+by = x(a+b) + y(a+b) = (a+b)(x+y)$$.

Show question

Question

Can you list down the three steps for factoring polynomials through grouping?

1. Group the polynomial into two sets of two terms by splitting the first two terms and the last two terms of the equation.
2. Factor out the GCF from each group.
3. If we find a common polynomial between the two factorized groups, factor out the GCF again to obtain the complete factorized form.

Show question

Question

What does the Zero Product Property tell us?

If the product of $$a$$ and $$b$$ is zero, then either $$a$$ is zero or $$b$$ is zero, or possibly both are zero.

Show question

Question

What is a Special Product?

A special product is a product of two binomials that hold a predictable pattern.

Show question

Question

How do special products help us in solving polynomials?

Special products do not require lengthy workings when solving them. We can expand and factorize polynomials easily by recognizing the pattern each technique holds.

Show question

Question

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

Factoring out the greatest common factor.

Show question

Question

True or false: The grouping method is usually used to factor polynomials with three terms.

False.

Show question

Question

What does FOIL stand for?

First, Inner, Outer, Last.

Show question

Question

If you wanted to solve $$15x^5 -25x^3 = 0$$, what would be your first step?

Factoring out $$5x^3$$.

Show question

Question

What are the factors of $$4xy+8y-3x-6$$?

$$4y-3$$.

Show question

Question

What are the factors of $$2x^3 - 8x^2 + 6x$$?

$$x-3$$.

Show question

Question

What is the Remainder Theorem used for?

The Remainder Theorem is used to find the remainder of a polynomial when it is divided by a linear polynomial.

Show question

Question

What is the Factor Theorem used for?

The Factor Theorem is used to completely factor a polynomial into a product of n factors.

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Question

What is the difference between the Factor and Remainder Theorems?

The Factor Theorem relates the factors of a polynomial to its roots while the Remainder Theorem links the remainder of division by a linear polynomial with the value of a function at a point.

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Question

What algorithm do we use to prove the Remainder and Factor Theorem?

The Division Algorithm

Show question

Question

What method of division can we use to verify the results we obtain from the Factor Theorem? Choose any of which applies.

1. long division
2. synthetic division
3. all of the above

iii. all of the above

Show question

Question

"The Factor Theorem is used to completely factorize a given polynomial."

From this statement, what will this help us achieve?

Once we have completely factorized a polynomial using the Factor Theorem, we can easily find the solutions to the expression.

Show question

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