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# Factoring Simple Polynomials

## Introduction

The factoring of polynomials can be seen as a reverse operation to expanding brackets and is used in many different areas of mathematics and real-life situations. This article focuses on factoring simple polynomials involving the methods of greatest common factor, regrouping and using identities.

Factoring simple polynomials is typically introduced around 7th or 8th grade, making it suitable for students aged between 12 and 15 years. However, with the proper guidance and practice, even students as young as 8 or 9 years may begin to grasp the basics of the topic.

## Math Domain

This topic falls under the algebraic domain of mathematics. It builds upon and extends basic arithmetic and the understanding of numerical factors, leading to more complex algebraic expressions and equations.

## Applicable Common Core Standards

Factoring simple polynomials directly aligns with the following Common Core State Standards for Mathematics:

CCSS.Math.Content.6.EE.A.3: Apply the properties of operations to generate equivalent expressions.

CCSS.Math.Content.7.EE.B.4: Use variables to represent quantities in a real-world or mathematical problem and construct simple equations and inequalities to solve problems by reasoning about the quantities.

## Definition of the Topic

Polynomials are algebraic expressions containing variables and coefficients. They are expressed in the form of anxn + an-1xn-1 + … + a2x2 + a1x + a0. The ‘a’s represent coefficients, and ‘n’ represents the degree of the polynomial.

Factoring polynomials involves expressing the polynomial as the product of two or more simpler polynomials. For example, the polynomial x2 – 9 can be factored into (x – 3)(x + 3).

## Key Concepts

Factors: Factors are quantities that we multiply to get the original number or expression. In terms of polynomials, factors are simpler polynomials that multiply to give the original polynomial.

Zero-product property: According to this property, at least one of the two numbers must be 0 if the product of two numbers is zero.

Factoring by grouping: This is a method for factoring polynomials when they have four or more terms. The terms are grouped in pairs, and the greatest common factor (GCF) is factored out from each pair.

Factoring trinomials: This is the process of breaking down a trinomial into the product of two binomial expressions.

## Discussion with Illustrative Examples

Let’s start by briefly discussing what factoring is. Finding what we multiplied to get the given quantity is done through the factoring procedure. We frequently use numbers in this way. Here are some examples of factoring 20.

(10)(2)=20

(40)(½)=20

(-4)(-5)=20

(4)(5)=20

(2)(2)(5)=20

2, 2, and 5 are the prime factors of 20 since we can no longer factor these numbers.

The process for factoring polynomials is essentially the same. We identify the terms that were multiplied to create the given polynomial.

The following are some of the methods of factoring polynomials.

Greatest Common Factor

Factoring by Regrouping Terms

Factoring Using Special Identities

Factoring Quadratic Polynomials of the Form  x2+(a+b) x+ab

### Greatest Common Factor

When factoring a polynomial with any number of terms, it is best to begin by determining whether there is a GCF—or greatest common factor—that all of the terms have.

Suppose we have 12x2 + 4x3 as the polynomial to be factored out. First, we shall get the polynomial’s greatest common factor (GCF). For the constants 12 and 4, we have the factors:

12 = 2×2×3

4=2×2

The GCF is 2×2=4.

For the variables x2 and x3, we have  their factors as:

x2=  x×x

x3=x×x×x

The GCF is x×x=x2

Multiplying the two GCFs then, we will get: 4x²

We can factor out 4x² from the polynomial where:

12x2= 3×4x2, and

4x3=x×4x2

Note that the operation in the example is in addition, so we write first the common factor, which is 4x²; then, we should have the other factors as the ones inside the groupings having the operation included (addition).

We now have our answer, 4x²(3 + x).

Therefore, 12x² + 4x³ = 4x² (3 + x).

### Factoring by Regrouping Terms

This technique is not used frequently, but it can be helpful when it is.

Suppose we have the expression 4mn+4m+3n+3. Notice that there is no common factor to all the terms, but 4mn+4m have common factors 4 and m while 3n+3 have a common factor which is 3.

4mn+4m=(4×m×n)+(4×m)

4mn+4m=(4×m)[n+(1)]

4mn+4m=(4m)(n+1)

3n+3=(3×n)+3

3n+3=3 (n+1)

Therefore, we have the common factor (n+1).

4mn+4m+3n+3=(4m)(n+1)+3 (n+1)

4mn+4m+3n+3=(n+1)(4m+3)

Hence, the factors are (n+1) and (4m+3).

Although factoring by grouping can be good, it only sometimes works.

### Factoring Using Special Identities

We can use the following common identities to factorize a given polynomial.

a2+2ab+b2=(a+b)2

a2-2ab+b2=(a-b)2

a2-b2=(a+b)(a-b)

a3+b3=(a+b)(a2-ab+b2)

a3-b3=(a-b)(a2+ab+b2)

Let us say we have the expression a2+10a+25. Notice that both a2 and 25 are perfect squares, and the middle term is positive of two times the product of the square root of a2 and 25. Hence, it fits the identity (a+b)2=a2+2ab+b2 where a=a and b=5.

a2+10a+25=a2+(2)(a)(5)+52

a2+10a+25=(a+5)2

Factoring Quadratic Polynomials of the Form  x2+(a+b)x+ab

There are polynomials with different forms than previously mentioned identities. For example, x2+6x+5, m2+9m+14, and t2+10d+16, etc. To factorize these expressions, we follow:

x2+(a+b)x+ab=(x+a)(x+b)

Let us say we want to find the factors of x2+13x+36. Here, ab=36 and a+b=13.

To find the values of a and b, we must consider the factors of 36 when added, which shall result in 13.

So we have 9×4=36 and 9+4=13.

x2+13x+36=x2+(9+4)x+36

x2+13x+36=x2+9x+4x+36

x2+13x+36=(x2+9x)+(4x+36)

x2+13x+36=(x)(x+9)+4(x+9)

x2+13x+36=(x+9)(x+4)

Hence, the factors are (x+9) and (x+4).

## Examples with Solutions

Example 1

Factor the polynomial x2 – 9.

Solution

Since both terms are squares but the second term is negative, we can express x2 – 9 as a difference of two squares that is (x)2-(3)2.

The identity a2-b2=(a+b)(a-b) is applicable in this expression where a=x and b=3. So we have,

x2-9=(x)2-(3)2

x2-9=(x+3)(x-3)

Therefore, the factors of x2-9 are (x+3) and (x-3).

Example 2

Factor the polynomial x2 + 5x + 6.

Solution

The given polynomial does not fit any of the identities but can be factored using the form

x2+(a+b)x+ab=(x+a)(x+b)

We are looking for two numbers (a and b) that, when multiplied, give 6 (the third term) and, when added, give 5 (the coefficient of the middle term). The numbers 2 and 3 meet this criterion.

So, we can factor the polynomial into (x + 2)(x + 3).

Example 3

Factorize: x2-7x+18.

Solution

The polynomial x2+7x-18 must take the form:

x2+7x-18=(x+___)(x+___)

Now, we must consider the factors of -18 that, when added, shall result in 7. Pair 9 and -2 meet the requirement. Hence,

x2+7x-18=(x+9)(x-2)

Example 4

Find the factors of x3+125.

Solution

The given polynomial fits the identity a3+b3=(a+b)(a2-ab+b2). Thus, we may write the expression as x3+53 where a=x and b=5.

So, we have,

x3-53=(x+5)(x2-(x)(5)+52)

x3-125=(x+5)(x2-5x+25)

## Real-life Application with Solution

Factoring polynomials is not just a mathematical exercise but has practical applications.

For example, you plan to build a rectangular garden with an area of 96ft2. The width of the garden is 4 feet less than the length. Find the dimensions of the garden.

Solution

Let x be the length of the garden, and x-4 be the width. Using the formula for the area of a rectangle (length×width), we have,

x (x-4) = 96

x2-4x=96 Distributive Property

x2-4x-96=0 Subtraction Property of Equality

(x-12)(x+8)=0 Factoring

Solving the resulting equation

x-12=0 ; x=12

x+8=0 ;x=-8

Since measurements cannot be negative, the length is 12 feet, and the width is 12-4=8 feet.

## Practice Test

Let’s test your understanding with these practice problems:

1. Factor the polynomial x2 – 16.

2. Factor the polynomial x2 + 7x + 10.

3. Factor the polynomial 2x2 – 8.

4. Factor the polynomial x2 – 5x + 6.

5. Factor the polynomial x2 + 10x + 21.

6. Factor the polynomial 3x2 – 15x + 18.

1. x2-16=(x+4)(x-4)

2. x2+7x+10=(x+5)(x+2)

3. 2x2-8=2(x2-4)

4. x2 – 5x + 6=(x-6)(x+1)

5. x2 + 10x + 21=(x+7)(x+3)

6. 3x2 – 15x + 18=3(x2-5x+6)=3(x-3)(x-2)

### Why do we need to factor polynomials?

Factoring polynomials helps to simplify complex expressions and solve equations. It is also essential for optimizing solutions in real-world scenarios.

### What is the zero-product property?

The zero-product property states that at least one of the two numbers must be 0 if the product of two numbers is zero. The zero-product property is very useful in solving equations.

### What is a binomial and a trinomial?

A binomial is a polynomial with two terms, while a trinomial is a polynomial with three terms.

### What is the difference between a coefficient and a variable?

In a polynomial, a variable is a symbol (often a letter) that can take on any number value, while a coefficient is a number that multiplies the variable.

### Can all polynomials be factored in?

Not all polynomials can be factored using integers and rational numbers. Some polynomials can only be factored using irrational numbers or complex numbers.

Remember, practice is vital when mastering the skill of factoring simple polynomials. So, keep practising and have fun exploring the world of algebra!