Factorization of Polynomial

The process of finding factors of a given value or mathematical expression is called factorization. Factors are the integers that are multiplied to produce an original number. For example, the factors of 18 are 2, 3, 6, 9 and 18, such as;

18 = 2 x 9

18 = 2 x 3 x 3

18 = 3 x 6

Similarly, in the case of polynomials, the factors are the polynomials which are multiplied to produce the original polynomial. For example, the factors of x2 + 5x + 6 is (x + 2) (x + 3). When we multiply both x +2 and x+3, then the original polynomial is generated. After factorization, we can also find the zeros of the polynomials. In this case, zeroes are x = -2 and x = -3.

Types of Factoring polynomials

There are six different methods for factorizing polynomials. The six methods are as follows:

  • Greatest Common Factor (GCF)
  • Grouping Method
  • Sum or difference in two cubes
  • Difference in two squares method
  • General trinomials
  • Trinomial method

In this article, let us discuss the two basic methods which we are using frequently to factorize the polynomial. Those two methods are the greatest common factor method and the grouping method. Apart from these methods, we can factorize the polynomials by the use of general algebraic identities. Similarly, if the polynomial is of a quadratic expression, we can use the quadratic equation to find the roots/factor of a given expression. The formula to find the factors of the quadratic expression (ax2+bx+c) is given by:

How to Solve Polynomials?

There are a certain number of methods by which we can solve polynomials. Let us discuss these methods.

Greatest Common Factor

We have to find out the greatest common factor, of the given polynomial to factorize it. This process is nothing but a type of reverse procedure of distributive law, such as;

p( q + r) = pq + pr

But in the case of factorization, it is just an inverse process;

pq + pr = p(q + r)

where p is the greatest common factor.

Factoring Polynomials By Grouping

This method is also said to be factoring by pairs. Here, the given polynomial is distributed in pairs or grouped in pairs to find the zeros.  Let us take an example.

Example: Factorise x2-15x+50

Find the two numbers which when added gives -15 and when multiplied gives 50.

So, -5 and -10 are the two numbers, such that;

(-5) + (-10) = -15

(-5) x (-10) = 50

Hence, we can write the given polynomial as;

x2-5x-10x+50

x(x-5)-10(x-5)

Taking x – 5 as common factor we get;

(x-5)(x-10)

Hence, the factors are (x – 5) and (x – 10).

Factoring Using Identities

The factorisation can be done also by using algebraic identities. The most common identities used in terms of the factorisation are:

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

Let us see an example:

Factorise (x2 – 112)

Using the identity, we can write the above polynomial as;

(x+11) (x-11)

Factor theorem

For a polynomial p(x) of degree greater than or equal to one,

  1. x-a is a factor of p(x), if p(a) = 0
  2. If p(a) = 0, then x-a is a factor of p(x)

Where ‘a’ is a real number.

Factoring Polynomial with Four Terms

Let us learn how to factorize the polynomial having four terms. For example, x3 + x2 – x – 1 is the polynomial. 

Break the given polynomial into two parts first.

(x3 + x2)+( –x – 1)

Now find the highest common factor from both the parts and take that factor out of the bracket. 

We can see, from the first part, x2 is the greatest common factor and from the second part we can take out the minus sign. Thus,

x2(x+1)-1(x+1)

Again, regrouping the terms as the factors.

(x2-1) (x+1)Therefore, the factorisation of x3+ x2 – x – 1 gives (x2 -1) (x+1)

Solved Examples

Question 1:

Check whether x+3 is a factor of x3 + 3x2 + 5x +15.

Solution:

Let x + 3= 0

=> x = -3

Now, p(x) = x3 + 3x2 + 5x +15

Let us check the value of this polynomial for x = -3.

p(-3) = (-3)3 + 3 (-3)2 + 5(-3) + 15 = -27 + 27 – 15 + 15 = 0

As, p(-3) = 0,  x+3 is a factor of x3 + 3x2 + 5x +15.

Factoring By Splitting the Middle Term

Question 2:

Factorize x2 + 5x + 6.

Solution:

Let us try factorizing this polynomial using splitting the middle term method.

Factoring polynomials by splitting the middle term:

In this technique we need to find two numbers ‘a’ and ‘b’ such that a + b =5 and ab = 6.

On solving this we obtain, a = 3 and b = 2

Thus, the above expression can be written as:

x2 + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2)

Thus, x+3 and x+2 are the factors of the polynomial x2 + 5x + 6.

Factorisation of Polynomial

The process of finding factors of a given value or mathematical expression is called factorisation. Factors are the integers that are multiplied to produce an original number. For example, the factors of 18 are 2, 3, 6, 9 and 18, such as;

18 = 2 x 9

18 = 2 x 3 x 3

18 = 3 x 6

Similarly, in the case of polynomials, the factors are the polynomials which are multiplied to produce the original polynomial. For example, the factors of x2 + 5x + 6 is (x + 2) (x + 3). When we multiply both x +2 and x+3, then the original polynomial is generated. After factorisation, we can also find the zeros of the polynomials. In this case, zeroes are x = -2 and x = -3.

Types of Factoring polynomials

There are six different methods to factorising polynomials. The six methods are as follows:

  • Greatest Common Factor (GCF)
  • Grouping Method
  • Sum or difference in two cubes
  • Difference in two squares method
  • General trinomials
  • Trinomial method

In this article, let us discuss the two basic methods which we are using frequently to factorise the polynomial. Those two methods are the greatest common factor method and the grouping method. Apart from these methods, we can factorise the polynomials by the use of general algebraic identities. Similarly, if the polynomial is of a quadratic expression, we can use the quadratic equation to find the roots/factor of a given expression. The formula to find the factors of the quadratic expression (ax2+bx+c) is given by:

How to Solve Polynomials?

There are a certain number of methods by which we can solve polynomials. Let us discuss these methods.

Greatest Common Factor

We have to find out the greatest common factor, of the given polynomial to factorise it. This process is nothing but a type of reverse procedure of distributive law, such as;

p( q + r) = pq + pr

But in the case of factorisation, it is just an inverse process;

pq + pr = p(q + r)

where p is the greatest common factor.

Factoring Polynomials By Grouping

This method is also said to be factoring by pairs. Here, the given polynomial is distributed in pairs or grouped in pairs to find the zeros.  Let us take an example.

Example: Factorise x2-15x+50

Find the two numbers which when added gives -15 and when multiplied gives 50.

So, -5 and -10 are the two numbers, such that;

(-5) + (-10) = -15

(-5) x (-10) = 50

Hence, we can write the given polynomial as;

x2-5x-10x+50

x(x-5)-10(x-5)

Taking x – 5 as common factor we get;

(x-5)(x-10)

Hence, the factors are (x – 5) and (x – 10).

Factoring Using Identities

The factorisation can be done also by using algebraic identities. The most common identities used in terms of the factorisation are:

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

Let us see an example:

Factorise (x2 – 112)

Using the identity, we can write the above polynomial as;

(x+11) (x-11)

Factor theorem

For a polynomial p(x) of degree greater than or equal to one,

  1. x-a is a factor of p(x), if p(a) = 0
  2. If p(a) = 0, then x-a is a factor of p(x)

Where ‘a’ is a real number.

Factoring Polynomial with Four Terms

Let us learn how to factorize the polynomial having four terms. For example, x3 + x2 – x – 1 is the polynomial. 

Break the given polynomial into two parts first.

(x3 + x2)+( –x – 1)

Now find the highest common factor from both the parts and take that factor out of the bracket. 

We can see, from the first part, x2 is the greatest common factor and from the second part we can take out the minus sign. Thus,

x2(x+1)-1(x+1)

Again, regrouping the terms as the factors.

(x2-1) (x+1)Therefore, the factorisation of x3+ x2 – x – 1 gives (x2 -1) (x+1)

Solved Examples

Question 1:

Check whether x+3 is a factor of x3 + 3x2 + 5x +15.

Solution:

Let x + 3= 0

=> x = -3

Now, p(x) = x3 + 3x2 + 5x +15

Let us check the value of this polynomial for x = -3.

p(-3) = (-3)3 + 3 (-3)2 + 5(-3) + 15 = -27 + 27 – 15 + 15 = 0

As, p(-3) = 0,  x+3 is a factor of x3 + 3x2 + 5x +15.

Factoring By Splitting the Middle Term

Question 2:

Factorize x2 + 5x + 6.

Solution:

Let us try factorizing this polynomial using splitting the middle term method.

Factoring polynomials by splitting the middle term:

In this technique we need to find two numbers ‘a’ and ‘b’ such that a + b =5 and ab = 6.

On solving this we obtain, a = 3 and b = 2

Thus, the above expression can be written as:

x2 + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2)

Thus, x+3 and x+2 are the factors of the polynomial x2 + 5x + 6.