In this chapter we will learn to factorize polynomial when it have common binomial function in each of its component.

Before moving on to the main topic, let us revise the basics first.

## Factorization and Factor of polynomial

The breaking of polynomial into smaller component is called factorization.

The smaller components produced after factorization are called factors of polynomial.

All the produced factors can completely divide the polynomial, leaving remainder 0.

So in this topic, we will try to factorize the polynomial when it has same binomial function in each of its components.

## How to factorize polynomial with common binomials ?

To Factorize the polynomial, follow the below steps;

(a) **Check if there is common binomial term** present in all entities of polynomial.

(b) **Separate the common binomial and simplify** the remaining expression.

(c) After the simplification, we will get the factors of given polynomial.

These three steps are easy and straight forward.

Let us understand the concept with the help of examples.**Example 01**

Factorize the below polynomial.

\mathtt{\Longrightarrow \ 2y( x+3) \ +\ 7( x+3)} **Solution**

The above polynomial consist of two entities separated by addition sign.

First entity ⟹ 2y ( x + 3 )

Second entity ⟹ 7 ( x + 3 )

Note that both entity have common binomial (x + 3).

Separating the common binomial and solving rest of the expression.

\mathtt{\Longrightarrow \ 2y( x+3) \ +\ 7( x+3)}\\\ \\ \mathtt{\Longrightarrow \ ( x+3) \ ( 2y+7)}

Hence, (x+3) & (2y+7) are the factors of given polynomial.

**Example 02**

Factorize the polynomial.

\mathtt{\Longrightarrow \ ( 2x-y)^{2} -11( 2x-y)}

**Solution**

The above polynomial contains two entity separated by subtraction sign.**First entity** ⟹ \mathtt{( 2x-y)^{2}} **Second entity** ⟹ 11 (2x – y)

Note that** (2x – y) is the common binomial** in both the entities.

Separating the common binomial and then solving rest of the expression.

\mathtt{\Longrightarrow \ ( 2x-y)^{2} -11( 2x-y)}\\\ \\ \mathtt{\Longrightarrow \ ( 2x-y)( 2x-y) \ -\ 11( 2x-y)}\\\ \\ \mathtt{\Longrightarrow \ ( 2x-y) \ \ ( \ ( 2x-y) \ -\ 11\ )}\\\ \\ \mathtt{\Longrightarrow \ ( 2x-y) \ ( 2x\ -y\ -11)}

Hence, **(2x – y) (2x – y -11) is the solution.**

**Example 03**

Factorize the below polynomial.

\mathtt{\ \left( x^{2} -y\right)( 2x+y) +\left( x^{2} -y\right)( 5x-7y)}

**Solution**

The polynomial consists of two entities separated by addition sign.**First entity** ⟹ \mathtt{\left( x^{2} -y\right)( 2x+y)} **Second entity** ⟹ \mathtt{\left( x^{2} -y\right)( 5x-7y)}

Note that \mathtt{\ \left( x^{2} -y\right)} is the common binomials among the given entities.

Separating the common term and solving rest of the expression.

\mathtt{\Longrightarrow \ \left( x^{2} -y\right)( 2x+y) +\left( x^{2} -y\right)( 5x-7y)}\\\ \\ \mathtt{\Longrightarrow \left( x^{2} -y\right)( 2x+y\ +5x\ -\ 7y)}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} -y\right) \ ( 7x\ -6y)}

Hence, \mathtt{\left( x^{2} -y\right)} and (7x – 6y) are the factors of given polynomial.

**Example 04**

Factorize the below polynomial.

\mathtt{\Longrightarrow \ 3\ ( x+5) \ +\ 7\ ( x+5) \ -\ 9\ ( x+5)}

**Solution**

Here the polynomial contains three terms.**First term** ⟹ 3 (x + 5)**Second term** ⟹ 7 (x + 5)**Third term** ⟹ 9 ( x + 5 )

Note that (x + 5) is the common binomial in all the above entities.

Separating the common binomial and solving rest of the expression, we get;

\mathtt{\Longrightarrow \ 3\ ( x+5) \ +\ 7\ ( x+5) \ -\ 9\ ( x+5)}\\\ \\ \mathtt{\Longrightarrow \ ( x\ +5) \ ( 3\ +\ 7\ -\ 9)}\\\ \\ \mathtt{\Longrightarrow \ ( \ x+\ 5) \ ( 1)}\\\ \\ \mathtt{\Longrightarrow \ ( x\ +\ 5)}

Hence, **( x + 5 ) is the factor of given polynomial**.

**Example 05**

Factorize the below polynomial.

\mathtt{\Longrightarrow \ 11( 2x+3) \ +\ 15\ ( 2x+3) \ -\ 2\ ( 2x+1)} **Solution**

The polynomial contain three entities.**First entity **⟹ 11 (2x + 3)**Second entity** ⟹ 15 (2x + 3)

**Third entity **⟹ 2 (2x + 1)

Note that there is no binomial which is common among all the three terms. So we cannot factorize, the polynomial fully with this method.

However, **(2x+3) is common in first and second term**. Let’s simplify the polynomial of the basis of this information.

\mathtt{\Longrightarrow \ 11( 2x+3) \ +\ 15\ ( 2x+3) \ -\ 2\ ( 2x+1)}\\\ \\ \mathtt{\Longrightarrow \ ( 2x+3) \ ( 11+15) \ -\ 2\ ( 2x+1)}\\\ \\ \mathtt{\Longrightarrow \ ( 2x+3) \ ( 26) \ -\ 2( 2x+1)}\\\ \\ \mathtt{\Longrightarrow \ 26\ ( 2x+3) \ -\ 2\ ( 2x+1)}

Note that **number 2 is common in both entity.**

\mathtt{\Longrightarrow \ 26\ ( 2x+3) \ -\ 2\ ( 2x+1)}\\\ \\ \mathtt{\Longrightarrow \ 2\ ( \ 13( 2x+3) \ -\ ( 2x\ +1) \ )}

To simplify further,** open up the brackets and solve**.

\mathtt{\Longrightarrow \ 2\ ( \ 13( 2x+3) \ -\ ( 2x\ +1) \ )}\\\ \\ \mathtt{\Longrightarrow \ 2\ ( \ 26x\ +39\ -2x-1\ )}\\\ \\ \mathtt{\Longrightarrow \ 2\ \ ( 24x\ +\ 38)}\\\ \\ \mathtt{\Longrightarrow \ 4\ \ ( \ 12x\ +\ 19\ )}

Hence, we got **4 and (12x + 9)** as the factors of given polynomial.