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# Factorization of Polynomial

Factorization of polynomials is required for solving various problems in Algebra. Factors are numbers or algebraic expressions that divide another number or expression evenly i.e their division yields a remainder 0, (or) factors are considered as small numbers or expressions which when multiplied give other numbers.

Example: 1, 2, 4, 7, 14, and 28 are factors of the number 28.

Similarly, a polynomial is factorized to write it as the product of two or more polynomials this is termed as factorization of polynomial

Example: ## Prime Factorization

The most common technique for finding the factors of any number or polynomial is prime factorization. In this method, we write a number in form of the product of its prime factors.

Example: Find the Prime Factors of 70

70 = 2 × 5 × 7
(here 2, 5, 7 are factors of 70; specially these are also called prime factors as these 2, 5, 7 are prime numbers)

Similarly, we can express algebraic expressions as the product of their factors. If an algebraic expression cannot be reduced further then its factors are called the prime factor.

Example:

8xy = 2 × 2 × 2 × x × y
(here 8xy is formed by multiplication of numbers(2, 2, 2, x, y) they are factors of 8xy)

## Types of Factoring Polynomials

Factorization is nothing but writing a number as the product of smaller numbers. It is the decomposition of a number (or) mathematical objects into smaller or simpler numbers/objects. Factorization of different types of algebraic expressions is very useful for various purposes used in mathematics. Various methods of factorization are,

• Greatest Common Factor (GCF)
• Regrouping
• Factorization Using Identities

### Greatest Common Factor

The highest common factor between the two numbers is called the GCF (Greatest Common Factor). It is useful for factoring polynomials

Steps for finding GCF are:

• Step 1: First, split every term of algebraic expression into irreducible factors
• Step 2: Then find the common terms among them.
• Step 3: Now the product of common terms and the remaining terms give the required factor form.

Example: Factorise 3x + 18

Solution:

Step 1: First splitting every term into irreducible factors.

3x = 3 × x
18 = 2 × 3 × 3

Step 2: Next step to find the common term

3 is the only common term

Step 3: Now the product of common terms and remaining terms is 3(x + 6)

So 3(x + 6) is the required form.

### Regrouping

Sometimes the terms of the given expression should be arranged in suitable groups in such a way so that all the groups have a common factor, and then the common factor is taken out. In this way factoring of a polynomial is done.

Example : Factorise x2 + yz + xy + xz

Solution:

Here we don’t have a common term in all terms, so we are taking (x2 + xy) as one group and  (yz + xz) as another group.

Factors of (x2 + xy) = (x × x) + (x × y)
= x(x + y)

Factors of (yz + xz) = (y × z) + (x × z)
= z(x + y)

After combining them,

x2 + yz + xy + xz = x(x + y) + z(x + y)

Taking (x + y) as common we get,

x2 + yz + xy + xz = (x + y) (x + z)

### Factorization Using Identities

There are many standard algebraic identities that are used to factorize various polynomials. Some of them are given below:

1. (a + b)2 = a2 + 2ab + b2
2. (a – b)2 = a2 – 2ab + b2
3. a2 – b2 = (a + b) (a – b)

Example 1: Factorise x2 + 8x + 16

Solution:

This is in the form of (a + b)2 = a2 + 2ab + b2

x2 + 8x + 16 = x2 + 2 × x × 4 + 42
= (x + 4)2
= (x + 4) (x + 4)

Example 2: Factorise x2 – 6x + 9

Solution:

This is in the form of (a – b)2 = a2 – 2ab + b2

x2 – 6x + 9 = x2 – 2 × x × 3 + 32
= (x – 3)2
= (x – 3) (x – 3)

### Factors Using Splitig Terms

Factoring polynomials is mostly used for solving quadratic equations. The quadratic equation is factorized to reduce it to linear factors. The form of a quadratic equation used is x2 + (a + b)x + ab = 0, that is split into two factors (x + a)(x + b) = 0. Consider the quadratic polynomial of the form

x2 + (a + b)x + ab.
= x.x + ax + bx + ab
= x(x + a) + b(x + a)
= (x + a)(x + b)

In the above polynomial, middle term is split as the sum or difference of two terms, and the constant term is the product of these two factors. Then, common factors are taken by grouping the first and second terms and the third and fourth terms. Thus, a quadratic polynomial is expressed as the product of two factors.

For example: x2 + 5x + 6

Solution:

= x2 + 5x + 6
= x.x + (3 + 2)x + 3.2
= x.x + 3x + 2x + 3.2
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)

Thus, factoring polynomials is done by splitting the middle terms in a quadratic polynomial.

## Solved Examples on Factorization

Example 1: Factorise a2 – 20a + 100

Solution:

This in the form of (a – b)2 = a2 – 2ab + b2

a2 – 20a + 100 = a2 – 2 × a × 10 + 102

= (a – 10)2

= (a – 10) (a – 10)

Example 2: Factorise 25x2 – 49

Solution:

This is the form of  a2 – b2 = (a + b) (a – b)

25x2 – 49 = (5x)2 – 72

= (5x + 7) (5x – 7)

Example 3: Factorise 2xy + 3 + 2y + 3x

Solution:

2xy + 2y + 3x + 3         [rearranging terms to get common terms]

= 2y (x + 1) + 3(x + 1)

= (2y + 3) (x + 1)

## FAQs on Factorization

Question 1: What is factorization and write its example?

Factorization is breaking a larger number into smaller number numbers so that when multiplied together, they give the original number. For example, factorization of 15 is achieved by multiplying 3 by 5.
i.e. 15 = 5 × 3

Question 2: What are the main 4 types of factorization methods?

The four main types of factoring methods are

• Greatest Common Factor (GCF),
• Grouping Method,
• Difference in two Squares,
• Sum or Difference in Cubes.

Question 3: How to factorize a given Algebraic Expression?

For factorizing an algebraic expression we use the known algebraic identities. For example: By using the algebraic identity (x-a) 2 = x2 -2ax +a2. x2 – 8x + 16 is factorized as (x – 4)(x – 4)

Question 4: Why is factorization important?