Factoring Methods

Factoring Methods

In this mini-lesson, we will explore the world of factoring methods in math. You will get to learn about the way to factorize an expression, solving quadratic equations by factoring and other interesting facts around the topic. You can try your hand at solving a few interesting practice questions at the end of the page.

Now, look at the picture given below.

different ingredients are combined together to make such yummy ice-creams.

Hey! Don't start thinking which one out of these is your favorite flavor.

Think about the ingredients which are combined together to make such yummy ice-creams.

Similarly, in algebra, we mix a few expressions together to form a new expression.

You can't take out each ingredient from these ice-creams but you can factor all the terms out from the expression.

In this lesson, we will learn various factoring methods and the way to factor quadratic equations.

Lesson Plan

What Is Factoring?

The process of finding the factors is called factoring.

All numbers have factors.

For example, 7 and 3 are the factors of 21.

7 and 3 are factors of 21.

Algebraic expressions also have factors.

For example, \(x\) and \(x-2\) are the factors of \(x^2-2x\)

Factors of Algebraic expression

Here, \(x\times 2x\) is the factored form of \(x^2-2x\).

So, factors can be numbers, algebraic expressions, or algebraic variables.

Experiment with the following simulation to see the factors of a few expressions.


Methods of Factoring

We will discuss some systematic methods of factoring algebraic expressions.

Method of Common Factors

Consider a simple example: \(3x+9\)

By factorizing each term we get, \((3\times x)+(3 \times 3)\)

By distributive law, \(3x+9=(3\times x)+(3 \times 3)=3(x+3)\).

Factors of 3x+9

Method of Regrouping

Regrouping allows us to rearrange the terms of the expression that leads to factorization. 

Consider an expression: \(2ab+2b+7a+7\)

Notice that there is no single factor common to all the terms.

Let us write \(2ab+2b\) and \(7a+7\) in the factor form separately.

\[\begin{align}2ab+2b&=2b(a+1)\\7a+7&=7(a+1)\end{align}\]

So, \(2ab+2b+7a+7=2b(a+1)+7(a+1)\).

Now, we have a common factor \((a+1)\) on the right-hand side.

The given expression is factorized as \(2ab+2b+7a+7=2b(a+1)+7(a+1)=(2b+7)(a+1)\).

Factoring by Method of Regrouping

Method Using Algebraic Identities

Remember a few identities that are used for factorization.

\((a+b)^2=a^2+b^2+2ab\)
\((a-b)^2=a^2+b^2-2ab\)
\((a+b)(a-b)=a^2-b^2\)

Let's factorize \(4z^{2}-12z+9\)

Observe that \(4z^{2}=(2z)^{2}\), \(12z=2\times 3\times 2z\), and \(9=3^{2}\)

So, we can write \(4z^{2}-12z+9=(2z)^{2}-2\times 2z\times 3+3^{2}=(2z-3)^2\)

Factoring by Method Using Algebraic Identities


How to Factor Trinomials?

Solving Quadratic Equations By Factoring

We’ll do a few examples on solving quadratic equations by factorization.

Example 1: \[4x-12x^2=0\]

Given any quadratic equation, first check for the common factors.

In this example, check for the common factors among \(4x\) and \(12x^2\)

Quadratic Equation: How to solve

We can observe that \(4x\) is a common factor. Let’s take that common factor from the quadratic equation.

\[\begin{align}4x-12x^2&=0\\4x(1-3x)&=0\end{align}\]

Thus, by factoring the quadratic equation, we get \(4x(1-3x)=0\).

In other words, the quadratic equation can be factored as \((4x)(1-3x)=0\).

Example 2: \[x^2+5x+6=0\]

We’ll factor this in two binomials, namely \((x+a)\) and \((x+b)\).

Expand \((x+a)(x+b)=0\).

\[\begin{align}(x+a)(x+b)&=0\\x(x+b)+a(x+b)&=0\\x^2+bx+ax+ab&=0\\x^2+(a+b)x+ab&=0\end{align}\]  

Let’s try to find \(a\) and \(b\) such that \((a+b)\) is mapped to 5 and \(ab\) is mapped to 6. 

Example showing Standard form of Quadratic Equation

  

The factors of \(6\) are \(1\), \(2\), \(3\) and \(6\). Find the pairs \(a\) and \(b\) from \(1\), \(2\), \(3\) and \(6\) such that \(a+b=5\) and \(ab=6\)

This is the right pair: \(2\) and \(3\), because, \(2+3 = 5\) and \(2 \times 3=6\)

Thus, the quadratic equation can be factored as \((x+2)(x+3)=0\).

Calculator For Factoring

Here is a calculator for factoring different expressions.

 
important notes to remember
Important Notes
  1. The process of finding the factors is called factoring.

  2. Factoring helps us to find the solution of any algebraic expression.

  3. Factoring allows us to express an expression in a simpler form.

Solved Examples

Example 1

 

 

George has a huge rectangular farm.

He knows that the area of the farm is \(528\;\text{ft}^{2}\)

Area of square farm is 528 square units.

He says "The length of the farm is one more than twice its breadth."

Can you use this information to write the factored form of 528?

Solution

Let the breadth of the farm be \(x\) feet.

So, the length of the farm would be \(2x+1\) feet.

The area of a rectangle is the product of its length and breadth.

According to the question,

\[\begin{align}\text{Area of Farm}&=\text{Breadth} \times \text{Length}\\528&=x\times (2x+1)\end{align}\]

So, the factored form of 528 is \(x(2x+1)\)
Example 2

 

 

Jenny asked Jolly to factorize \(6xy-4y+6-9x\)

Jolly wants to factorize it using the method of regrouping.

Jenny and Jolly are doing factorization

Can you help them?

Solution

We observe that we have no common factor among all the terms in the expression \(6xy-4y+6-9x\)

Let's deal with \(6xy-4y\) and \(6-9x\) separately.

\[\begin{align}6xy-4y&=2y(3x-2)\\6-9x&=-3(3x-2)\end{align}\]

So, the given expression can be written as 

\[\begin{align}6xy-4y+6-9x&=2y(3x-2)-3(3x-2)\\&=(2y-3)(3x-2)\end{align}\]

So, factors of \(6xy-4y+6-9x\) are \((2y-3)\) and \((3x-2)\)
Example 3

 

 

Mia is a fitness enthusiast who runs every morning.

The park where she jogs is rectangular in shape and measures 12 feet by 8 feet.

Quadratic Equation Question to find width of a park's pathway

An environmentalist group plans to revamp the park and decides to build a pathway surrounding the park.

This would increase the total area to 140 sq. ft.

What will be the width of the pathway?

Solution

Let’s denote the width of the pathway as \(x\).

Then, the length and breadth of the outer rectangle is \((12+2x)\;\text{ft.}\) and \((8+2x)\;\text{ft.}\)

According to the question,
\[\begin{align}(12+2x)(8+2x)&=140\\2(6+x)\cdot 2(4+x)&=140\\(x+6)(x+4)&=35\\x^2+10x-11&=0\end{align}\]

So, we have to find the solution of the quadratic equation \(x^2+10x-11=0\)

\[\begin{align}x^2+11x-x-11&=0\\x(x+11)-(x+11)&=0\\(x+11)(x-1)&=0\\x&=1,-11\end{align}\]

Since the length can’t be negative, we take \(x=1\)

So, the width of the pathway will be 1 feet.

Sometimes working with a hard expression is difficult. But don’t worry! 

Here are a few tips and tricks for you to follow while factoring.

 
tips and tricks
Tips and Tricks
  1. Check out for any common terms in an expression and take the greatest common factor.
  2. Check if any algebraic identities are applicable in the expression.
  3. Keep factoring the expression until you reach the simplest form, that is, the form that is not further divisible.

Interactive Questions

Here are a few activities for you to practice.

Select/type your answer and click the "Check Answer" button to see the result.

 
 
 
 
 

Let's Summarize

The mini-lesson targeted the fascinating concept of factoring methods. The math journey around factoring methods starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

About Cuemath

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.


Frequently Asked Questions (FAQs)

1. How to factor trinomials with a leading coefficient?

The following steps show how to factor trinomials with a leading coefficient.

  1. Write the trinomial in the standard form.
  2. Take the greatest common factor out if it exists.
  3. Find the product of the leading coefficient and the constant term.
  4. Determine the factors of the product found in step 3 and check which factor pair will result in the coefficient of \(x\).
  5. After choosing the appropriate factor pair, keep the sign in each number such that while operating them we get the result as the coefficient of \(x\) and on finding their product the number is equal to the number found in step 3.
  6. Now, you have 4 terms in the expression and so we use the method of regrouping to factorize.

Flow chart of factor trinomials with a leading coefficient

2. How to factor perfect square trinomials?

Use the algebraic identities mentioned below to factor perfect square trinomials.

  1. \((a+b)^2=a^2+b^2+2ab\)
  2. \((a-b)^2=a^2+b^2-2ab\)
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