Factorization of Polynomial

Factorization in mathematics is the process of expressing a number or an algebraic expression as a product of simpler factors. Factorizing a polynomial means expressing it as a product of simpler polynomials or monomials. This helps simplify expressions, solve equations, and understand polynomial behavior.

Example: The factors of 12 are 1, 2, 3, 4, 6, and 12, and we can express 12 as 12 = 1 × 12, 2 × 6, or 4 × 3.

Prime Factorization of Polynomials

Just like numbers, algebraic expressions can also be factored into simpler polynomials. If an algebraic expression cannot be simplified further, then its factors are considered to be prime factors of the expression.

Example: Consider the polynomial 8xy.

The prime factorization of 8xy is:

8xy = 2 × 2 × 2 × x × y

Here, 2, x, and y are the prime factors of 8xy, considering that x and y are relatively prime to each other (they don't share any common factors other than 1).

Methods of Factorization of Polynomials

Factorization is the process of 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 of polynomials are

• By Greatest Common Factor (GCF),
• By Regrouping,
• Factorization Using Identities,
• Factorization Using Splitting Terms (only for quadratic polynomials),
• Factorization Using Rational Roots and Factor Theorem.

Factorization of Polynomials by Greatest Common Factor

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

The 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.

Factorization of Polynomials by 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)

Thus, (x + y), and (x + z) are the factors of the given expression x² + yz + xy + xz

Factorization of Polynomials Using Identities

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

• a² + 2ab + b² = (a + b)²
• a² - 2ab + b² = (a - b)² 
• a² - b² = (a + b) (a - b)
• a² + b² + c² + 2(ab + bc + ca) = (a + b + c)² 
• a³ + b³ + 3ab(a + b) = (a + b)³
• a³ - b³ - 3ab(a - b) = (a - b)³
• a³+ b³= (a+b)(a²+ b²– ab)
• a³ - b³= (a-b)(a²+ b²+ ab)
• a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

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:

Let g(x) = x² - 6x + 9

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

⇒ g(x) = x² - 2 × x × 3 + 3²
⇒ g(x) = (x - 3)²
⇒ g(x) =(x - 3) (x - 3)

Example 3: Factorise x³ - 27.

Solution:

Let g(x) = x³ - 27

Using, a³ - b³= (a-b)(a²+ b²+ ab)

g(x) = x³ - 3³ 
⇒ g(x) = (x- 3 )(x² + 3² + 3x) 
⇒ g(x) = (x- 3 )(x²+ 3x +  9 ) 

As , x²+ 3x +  9 can't be factorize further,

Thus factors of x³ - 27 are  (x- 3 ), and (x²+ 3x +  9 ) .

Related Articles:

• Zeros of Polynomial 
• Factoring Polynomials
• Factor Theorem

Examples on Factorization

Example 1: Factorize a² - 20a + 100

Solution:

f(a) = a² - 20a + 100 

This is the form of (a - b)² = a² - 2ab + b²
⇒ f(a) = a² - 2 × a × 10 + 10²
⇒ f(a)  = (a - 10)²
⇒ f(a)  = (a - 10) (a - 10)

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

Solution:

Let f(x) =  2xy + 3 + 2y + 3x

⇒ f(x) = 2xy + 2y + 3x + 3         [rearranging terms to get common terms]
⇒ f(x) = 2y (x + 1) + 3(x + 1)
⇒ f(x) = (2y + 3)(x + 1)

Example 4: Factorize x³ - 8

Solution:

We know the identity, a³ - b³ = (a - b)(a² + ab + b²). Here a = x and b = 2.

Thus, x³ - 8= (x - 2)(x² + 2x + 2²)
⇒ x³ - 8= (x - 2)(x² + 2x + 4 )

Example 5: Factorize 2x³ + 4x²- 6x.

Solution:

Let f(x) = 2x³ + 4x²- 6x 

⇒ f(x) = 2x(x² + 2x - 3)
⇒ f(x) = 2x(x² + 3x - x - 3)
⇒ f(x) = 2x[x(x + 3) - (x + 3)]
⇒ f(x) = 2x(x + 3)(x - 1)

Example 6: Factorise a⁴ – 81

Solution:

Using the identity a² − b² = (a − b)(a + b), we get:
a⁴ − 81 = (a²)² − (9)² = (a² − 9)(a² + 9)
= (a − 3)(a + 3)(a² + 9).

Unsolved Question

Question 1: Factorize: x² + 14x + 49

Question 2: Factorize: 64y² − 100

Question 3: Factorize: 3ab + 6a + 2b + 4

Question 4: Factorize: x³ + 27

Question 5: Factorize: 5x³ − 20x

Question 6: Factorize: a⁴ − 16