Factorisation by Direct Method & Grouping: Definition, Examples & Notes

Factorisation by Direct Method and Grouping: Factorisation, also referred to as factoring, is the process of representing a number or any mathematical object as a product of several factors, typically smaller or simpler items of the same kind. In every case, the result is a product of simpler things. Factoring aims to reduce something to its “basic building blocks,” such as reducing polynomials to irreducible polynomials or reducing integers to prime numbers. The fundamental theorem of arithmetic covers factoring integers, while the fundamental theorem of algebra covers factoring polynomials.

There are different methods to factorise polynomials. In this article, we will learn about the factorisation of algebraic expressions using the method of taking common factors and factorising algebraic expressions using grouping terms.

Define Factorisation by Direct Method and Grouping

Direct method (Taking out common factors)

Take out the common factors and divide each term of the expression by the common factor when the polynomial terms contain common factors. The quotient should be written in brackets, whereas the common factor should be written outside the brackets. 

Example: \(30 x^{3} y^{2}+45 x^{2} y^{3}\)

Solution: Here, \(15 x^{2} y^{2}\) is the common factor.

\(\therefore 30 x^{3} y^{2}+45 x^{2} y^{3}=15 x^{2} y^{2}(2 x+3 y)\)

Grouping the Terms

A polynomial with common factors can be determined using this method by grouping its terms in the appropriate order. To factor a polynomial by grouping the terms, follow the procedures below.

i. Group the terms of the given algebraic expression such that each group has a common factor.

ii. Take out the common factors.

Example: In the expression \(ax+ay+bx+by,\) we can factorise the expression by rearranging the terms.

Solution: \(ax+ay+bx+by\)

\(=(a x+b x)+(a y+b y)\)  (Rearranging the terms)

\(=x(a+b)+y(a+b)\) [ Here \((a+b)\)  is a common factor.]

\(=(a+b)(x+y)\)

Thus, \((a+b)\) and \((x+y)\) are the factors of \((ax+ay+bx+by).\)

Examples of Factorisation by Direct Method and Grouping

Example for Factorisation by the Direct Method

Question: \(20\left(p^{2}-q^{2}\right)^{2}-30\left(p^{2}-q^{2}\right)^{3}+5\left(p^{2}-q^{2}\right)\)

Solution:  The given expressions is \(20\left(p^{2}-q^{2}\right)^{2}-30\left(p^{2}-q^{2}\right)^{3}+5\left(p^{2}-q^{2}\right)\)

\(=5\left(p^{2}-q^{2}\right)\left[4\left(p^{2}-q^{2}\right)-6\left(p^{2}-q^{2}\right)^{2}+1\right]\)

\(=5\left(p^{2}-q^{2}\right)\left[4\left(p^{2}-q^{2}\right)-6\left(p^{4}+q^{4}-2 p^{2} q^{2}\right)+1\right]\)

\(=5\left(p^{2}-q^{2}\right)\left(4 p^{2}-4 q^{2}-6 p^{4}-6 q^{4}+12 p^{2} q^{2}+1\right)\)

\(=5(p+q)(p-q)\left(4 p^{2}-4 q^{2}-6 p^{4}-6 q^{4}+12 p^{2} q^{2}+1\right)\)

Example for factorisation by grouping

Factorise the following: \(m^{2}-2 m(1+l)+4 l\)

Solution: The given expression is \(m^{2}-2 m(1+l)+4 l\)

\(=m^{2}-2 m-2 l m+4 l\)

\(=m^{2}-2 l m-2 m+4 l\)

\(=m(m-2 l)-2(m-2 l)\)

\(=(m-2 l)(m-2)\)

Hence, the factors of the expression \(m^{2}-2 m(1+l)+4 l\) is \((m-2 l)(m-2)\)

Factorisation by Direct Method and Grouping Notes

The factorisation is the process of expressing an algebraic expression as the product of two or more factors.

If the different terms of a given polynomial have common factors, the polynomial can be factored using the approach below:

i. Determine the HCF of all the polynomial terms. 

ii. Use HCF to divide each term of the given polynomial. Keep the common factor outside the brackets and enclose the quotient within them.

Example: Factorise the following polynomial: \(24 x^{3}-32 x^{2}\)

Solution: The given expression is \(24 x^{3}-32 x^{2}\)

Here, the HCF of \(24 x^{3}\) and \(32 x^{2}\) is \(8 x^{2}\)

Divide each term by \(8 x^{2}\) and keep \(8 x^{2}\) outside the bracket.

That is, 

\(24 x^{3}-32 x^{2}=8 x^{2}\left(\frac{24 x^{3}}{8 x^{2}}-\frac{32 x^{2}}{8 x^{2}}\right)\)

\(=8 x^{2}(3 x-4)\)

Hence, the factors of the expression \(24 x^{3}-32 x^{2}\) is \(8 x^{2}(3 x-4)\)

Factorising by the Grouping of Terms

When the grouping of the given polynomial gives rise to a common factor, then the given polynomial can be factorised by the following procedure:

1. Arrange the terms of the given polynomial in groups so that each group has a common factor.
2. Factorise each group.
3. Take out the factor which is common to each group.

Note: Factorisation by grouping is possible only if the given polynomial contains an even number of terms.

Example: Factorise the following polynomial: \(a b\left(x^{2}+y^{2}\right)+x y\left(a^{2}+b^{2}\right)\)

Solution: The given expression is \(a b\left(x^{2}+y^{2}\right)+x y\left(a^{2}+b^{2}\right)\)

\(=a b x^{2}+a b y^{2}+a^{2} x y+b^{2} x y\)

\(=\left(a b x^{2}+a^{2} x y\right)+\left(a b y^{2}+b^{2} x y\right)\)

\(=a x(b x+a y)+b y(a y+b x)\)

\(=a x(b x+a y)+b y(b x+a y)\)

\(=(b x+a y)(a x+b y)\)

Hence, the factors of the expression \(a b\left(x^{2}+y^{2}\right)+x y\left(a^{2}+b^{2}\right)\) is \((b x+a y)(a x+b y)\)

How to Calculate Factorisation by Direct Method and Grouping

Method of Common Factors or Direct Method

Taking common factors of each of the terms of the provided expression is the simplest way of factoring an algebraic equation. The factors of each of the terms of the algebraic expression are written as a first step. To obtain the possible factors, the common factors across the terms are also considered. This is the same as applying the distributive property backwards. With the help of an example, let us better understand this. 

Example: Factorise the following \(4 x+8\)

Solution: The given expression is \(4x+8\)

We can find the solution by factoring each term \(4×x+4×2\)

By distributive law, \(4x+8=4(x+2)\)

Hence, the factors of the expression \(4x+8\) is \((4x+2).\)

Factorisation by Grouping Method

The approach of factoring polynomials by grouping is a step further than identifying common factors. The goal here is to find groups from common factors to get the factors of a given polynomial equation. The polynomial expression’s number of terms is reduced to a smaller number of groups. To determine the group of factors, we first separate each term of the supplied expression into its factors and then look for common terms. With the help of the following example, let’s try to understand grouping for factorizing.

Example: Factorise the following \(4\,ab + 4\,b + 12\,a + 12.\)

Solution: The given expression is \(4ab+4b+12a+12\)

It’s important to note that \(4\) is a common factor in all the polynomial terms.

So we can write \(4 a b+4 b+12 a+12=4(a b+b+3 a+3)\)

\(=4[(a b+b)+(3 a+3)]\)

\(=4[b(a+1)+3(a+1)]\)

\(=4(a+1)(b+3)\)

Hence, the factors of the given expression \(4 a b+4 b+12 a+12\) is \(4(a+1)(b+3)\).

Solved Examples – Factorisation by Direct Method and Grouping

Q.1. Factorise the following expression \(xy-pq+qy-px.\)
Ans: The given expression is \(xy-pq+qy-px.\)
Since \(xy\) and \(pq\) have nothing in common, we do not group the terms in pairs in the order the given expression is written. Here we interchange \(-pq\) and \(-px.\)
That is, \(xy-pq+qy-px\)
\(=(x y-p x)+(q y-p q)\)
\(=x(y-p)+q(y-p)\)
\(=(y-p)(x+q)\)
Hence, the factors of the given expression \(x y-p q+q y-p x\) is \((y-p)(x+q)\)

Q.2. Factorise the following expression \(ax+by+bx+az+ay+bz.\)
Ans: The given expression is \(ax+by+bx+az+ay+bz\)
\(=(a x+a y+a z)+(b x+b y+b z)\)
\(=a(x+y+z)+b(x+y+z)\)
\(=(x+y+z)(a+b)\)
Hence, the factors of the given expression \(a x+b y+b x+a z+a y+b z\) is \((x+y+z)(a+b)\)

Q.3. Factorise the following expression \(12 x^{3} y^{4}+16 x^{2} y^{5}-4 x^{5} y^{2}.\)
Ans: The given expression is \(12 x^{3} y^{4}+16 x^{2} y^{5}-4 x^{5} y^{2}\)
The greatest common factor of the terms \(12 x^{3} y^{4}, 16 x^{2} y^{5}\) and \(4 x^{5} y^{2}\) of the expression \(12 x^{3} y^{4}+16 x^{2} y^{5}-4 x^{5} y^{2}\) is \(4 x^{2} y^{2}\)

Also, we can write \(12 x^{3} y^{4}=4 x^{2} y^{2} \times 3 x y^{2}, 16 x^{2} y^{5}=4 x^{2} y^{2} \times 4 y^{3}\) and, \(4 x^{5} y^{2}=4 x^{2} y^{2} \times x^{3}\)

Therefore, \(12 x^{3} y^{4}+16 x^{2} y^{5}-4 x^{5} y^{2}=4 x^{2} y^{2} \times 3 x y^{2}+4 x^{2} y^{2} \times 4 y^{3}-4 x^{2} y^{2} \times x^{3}\) 

\(=4 x^{2} y^{2}\left(3 x y^{2}+4 y^{3}-x^{3}\right)\)

Hence, the factors of the given expression \(12 x^{3} y^{4}+16 x^{2} y^{5}-4 x^{5} y^{2}\) is \(4 x^{2} y^{2}\left(3 x y^{2}+4 y^{3}-x^{3}\right)\)

Q.4. Factorise the following expression \((x+y)(2 a+b)-(3 x-2 y)(2 a+b)\).
Ans: The given expression is \((x+y)(2 a+b)-(3 x-2 y)(2 a+b)\)
\(={(x+y)-(3 x-2 y)}(2 a+b)\) [Taking \((2 a+b)\) common]
\(=(x+y-3 x+2 y)(2 a+b)\)
\(=(-2 x+3 y)(2 a+b)\)
Hence, the factors of the given expression \((x+y)(2 a+b)-(3 x-2 y)(2 a+b)\) is \((-2 x+3 y)(2 a+b)\)

Q.5. Factorise the following expression \(15xy-6x+10y-4.\)
Ans: We have, \(15xy-6x+10y-4\)
Here, none of the terms has anything in common. There are also four terms. So, imagine grouping the terms in pairs so that some factors are common to the terms in each pair, and after removing the common factors from each pair, the same binomial is left inside the two brackets. We can see that the first two terms share a common factor of \(3x.\) We have \(3x\) common by taking \(3x\) common from them, we get:
\(15 x y-6 x=3 x(5 y-2)\)
The last two terms have \(2\) as the common factor. Taking \(2\) common from these two, we get
\(10 y-4=2(5 y-2)\)
Clearly, \((5y-2)\) is the binomial common between these two groups. Thus, we group the terms as follows: \(15 x y-6 x+10 y-4=3 x(5 y-2)+2(5 y-2)\)
\(=(3 x+2)(5 y-2)\)
Hence, the factors of the given expression \(15 x y-6 x+10 y-4\) is \((3 x+2)(5 y-2)\)

Summary

In this article, we learnt about the definition of factorisation by direct method and grouping, examples of factorisation by direct method and grouping, factorisation by direct method and grouping notes, how to calculate factorisation by direct method and grouping, solved examples on factorisation by direct method and grouping and FAQs on factorisation by direct method and grouping.

Frequently Asked Questions (FAQs)- Factorisation by Direct Method and Grouping

Frequently asked questions related to factorisation by direct method and grouping is listed as follows:

Q.1. What is the grouping method of factoring?
Ans: The approach of factoring polynomials by grouping is a step further than identifying common factors. The goal here is to find groups from common factors to get the factors of a given polynomial equation. The polynomial expression’s number of terms is reduced to a smaller number of groups. To determine the group of factors, we first separate each term of the supplied expression into its factors and then look for common terms.

Q.2. What are the 4 types of factorisations?
Ans: The four types of factorisations are:
i. Taking out common factors
ii. Grouping
iii. Using algebraic identities
iv. Splitting the middle term 

Q.3. What is factorisation?
Ans: Factorisation is the process of expressing an algebraic expression as the product of two or more factors.

Q.4. How do you factor a polynomial by grouping?
Ans: 
Example: Factorise \(6 a b-2 a+9 b-3\),
We have, \(6ab-2a+9b-3\):
Here, none of the terms has anything in common. There are also four terms. So, imagine grouping the terms in pairs so that some factors are common to the terms in each pair, and after removing the common factors from each pair, the same binomial is left inside the two brackets. We can see that the first two terms share a common factor of \(2a.\) We have \(2a\) common by taking \(2a\) common from them, we get:
\(6 a b-2 a=2 a(3 b-1)\)

The last two terms have \(3\) as the common factor. Taking \(3\) common from these two, we get:
\(9 b-3=3(3 b-1)\)
Clearly, \((3b-1)\) is the binomial common between these two groups. Thus, we group the terms as follows: \(6 a b-2 a+9 b-3=2 a(3 b-1)+3(3 b-1)\)
\(=(3 b-1)(2 a+3)\)
Hence, the factors of the given expression \(6ab-2a+9b-3\) is \((3b-1)(2a+3).\)

Q.5. How do you factor common terms?
Ans: Taking common factors of each of the terms of the provided expression is the simplest way of factoring an algebraic equation. The factors of each of the terms of the algebraic expression are written as a first step. To obtain the possible factors, the common factors across the terms are also considered.

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