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Factorization of Polynomial

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Factorization is the process in which we can find factors of either a given number or the algebraic expression using various techniques such as prime factorization, factorization using algebraic identities, and factorization of a quadratic polynomial using splitting the middle term method. Factorization of polynomials or numbers can help us solve different problems in algebra. Factors are either numbers or expressions defined for given numbers and expressions such that when their division yields a remainder 0 or the number is less than the given number. For example, 1, 2, 4, 7, 14, and 28 are the factors of 28, and (x +1) and (x – 1) are the factors of f(x) = x2 – 1. The image of the factorization of the Polynomial is added below,

Factorization of polynomial

 

Prime Factorization Definition

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

Prime Factorization of Numbers

Prime factorization is the technique in which we can find the factors of numbers using division by the smallest prime possible at a time. i.e., if We need to find prime factors of 24, then first we’ll divide 24 by 2, which gives 0 as a remainder, and again divide the quotient 12 by 2(as it can be divided by 2), the remainder is still 0. Now again divide by 2 to get 6 as quotient, and again by 2 to get 3 (which is not divisible by 2). Thus divide it by the next prime which is 3 itself. Thus prime factorization of 24 = 2 × 2 × 2 × 2 × 3.

Example: Find the Prime Factors of 70

Solution:

As 70 = 2 × 5 × 7
(here 2, 5, 7 are prime factors of 70; as there are more factors such as 10, 14, 35 for 70 but these are not prime factors)

Prime Factorization of Polynomials

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, considering x and y relatively prime to everything.)

Factorization of 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 of polynomials are, 

  • By Greatest Common Factor (GCF)
  • By Regrouping
  • Factorization Using Identities
  • Factorization Using Splitig Terms (only for quadratic polynomials)
  • Factorization Using Rational Root 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 ) .

Factors Using Splitting 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, which is split into two factors (x + a)(x + b) = 0. Consider the quadratic polynomial of the form 

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

In the above polynomial, the 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) + 2(x + 3)
= (x + 3)(x + 2)

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

Factorization Using Rational Root and Factor Theorem

Factors of cubic or any higher polynomial can be found using the rational root theorem and long division together. Which is explained as follows:

Step 1: Use the rational root theorem to find the possible roots. i.e., For a polynomial of n degree i.e., f(x)  = anxn + an−1​xn−1 +…+ a1​x + a0, all the possible rational roots are the divisor of a0/an

Step 2: After finding one root, use the factor theorem to get the first factor and divide the polynomial with the factor using long division and write the polynomial as a product of quotient and dividend.

Step 3: If the quotient is a quadratic polynomial then solve it by the method of splitting the middle term. If not a quadratic polynomial then repeat step 1 and step 2 until the quotient becomes a quadratic polynomial.

Step 4: The result of step 3 is the required factors of the given polynomial.

Example: Factorise f(x) = x3 + 3x2 – 4x – 12.

Solution:

f(x) =x3 + 3x2 – 4x – 12

As p/q = -12

By rational root theorem, all possible rational roots of the polunomial are divisor of a0/an ,

Thus, divisors = ±1, ±2, ±3, ±4, ±6, and ±12.

x = 2, in p(x), we get

f(2) = (2)3 + 3(2)2 – 4(2) -12

f(2) = 8 + 12 – 8 – 12 = 0

Thus, by factor theorem, x – 2 is the factor of f(x).

Example of long division in Polynomial

 

Thus, x3 + 3x2 – 4x – 12= (x-2)(x2 +5x + 6)

x3 + 3x2 – 4x – 12 = (x-2)(x+2)(x+3)

Thus, factors of f(x) are (x-2), (x+2), and (x+3).

Also, Read

Solved Examples on Factorization

Example 1: Factorise a2 – 20a + 100

Solution:

f(a) = a2 – 20a + 100 

This in 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: 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:

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 – 83

Solution:

As we know, a2 – b2 = (a + b) (a – b)

Thus, a4 – 83 = a4 – 34 = (x² + 9) (x² – 9)

Now, again factorize (a2 – 9) using sam identity.

⇒ a4 – 83 = (a² + 9) (a – 3)(a + 3)

FAQs on Factorization

Q1: What is Factorization and write its example?

Answer:

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

Q2: What are the Methods of Factorization of Polynomials?

Answer:

The various different methods to factorize polynomials are as follows:

  • Greatest Common Factor (GCF)
  • Grouping Method
  • Factorization using Identities
  • Factors Using Splitting Terms
  • Factorization Using Rational Root and Factor Theorem

Q3: How to Factorize a Given Algebraic Expression?

Answer:

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)

Q4: What are the factors of 36?

Answer:

 The factors of 36 are 2, 3, 4, 6, 9, 12, and 18, i.e.

  • 2×18 = 36
  • 3×12 = 36
  • 4×9 = 36
  • 6×6= 36

Q5: What are the factors of 24?

Answer:

 The factors of 24 are 2, 3, 4, 6, 8, and 12, i.e.

  • 2×12 = 24
  • 3×8 = 24
  • 4×6 = 24

Q6: What are the factors of 18?

Answer:

 The factors of 18 are 2, 3, 6, and 9, i.e.

  • 2×9 = 18
  • 3×6 = 18

Q7: What are the factors of 12?

Answer:

 The factors of 12 are 2, 3, 4, and 6, i.e.

  • 2×6 = 12
  • 3×4 = 12

Q8: What are the factors of 30?

Answer:

 The factors of 30 are 2, 3, 5, 6, 10 and 15, i.e.

  • 2×15 = 30
  • 3×10 = 30
  • 5×6 = 30

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Last Updated : 09 Nov, 2023
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