Methods of Factorisation: Definition, Types and Examples

In Mathematics, factorisation or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler things of the same kind. In all cases, a product of simpler things is obtained. The aim of factoring is usually to reduce something to “basic building blocks”, such as numbers to prime numbers or polynomials to irreducible polynomials. Factoring integers is covered by the fundamental theorem of arithmetic and factoring polynomials by the fundamental theorem of algebra.

Ancient Greek mathematicians first considered factorisation in the case of integers. They proved the fundamental theorem of arithmetic, which asserts that each whole number could be factored into a product of prime numbers, that cannot be further factored into integers greater than \(1.\)

Methods of Factorisation: Definition

Prime factorisation: The process of stating a given number as the product of prime factors is called a prime factorisation or complete factorisation of the given number.

Example: Prime factorisation of \(36\) is \(2 \times 2 \times 3 \times 3\)

Factors: If an algebraic expression is written as the product of numbers or algebraic expressions, then each of these numbers and expressions is called the factors of the given algebraic expression, and the algebraic expression is called the product of these expressions.

Factorisation of algebraic expression: The process of writing a given algebraic expression as the product of two or more factors is called factorisation of algebraic expressions.

Example: Factorisation of \(3{x^2}y = 3 \times x \times x \times y\)

Methods of Factorisation Types

Prime factorisation of numbers

1. Prime factorisation of numbers using factor tree method
2. Prime factorisation of numbers using repeated division method

Factorisation of algebraic expression

1. Factorisation of algebraic expressions using the method of taking common factors
2. Factorisation of algebraic expressions using by regrouping terms
3. Factorisation of algebraic expressions using standard identities
4. Factorisation of quadratic expression using the method of splitting the middle term

Methods of Factorisation Notes

Prime factorisation of numbers using factor tree method: The factor tree method is quite flexible – at each branch, you can break the number into any factors until you reach the prime factors.

The following diagrams show the prime factorisation of numbers using the factor tree method.

Prime factorisation of \(36\) using factor tree method:

Prime factorisation of numbers using repeated division method: We will divide the number by the lowest prime number to find the prime factors. We will repeat this until the number is no longer divisible by that number. Then, we will divide it with the next higher prime number and repeat this process until the number is wholly factored with all the factors as prime numbers.

The following shows the prime factorisation of numbers using the repeated division method.

Prime factorisation of \(28\) using repeated division method:

Factorisation of algebraic expressions using the method of taking common factors

Find the factors of given terms.

Write the common factors in all the terms, putting a sign of multiplication between them.

Product of all common factors in all terms will be the required common factor.

Example: \(3{x^2} + 6xy \Rightarrow 3x(x + 2y)\)

Factorisation of expression in which only one factor (monomial) is common in each term

Working Rule:

Make groups of terms of given expression in such a way that each group has some common factor.

Write common factors in a bracket.

Divide expression in each group by common factor of that group and write the quotient in another bracket.

The sum of common factors of groups and the sum of quotients will be the required factors.

Factorisation of algebraic expressions using standard identities

Some expressions can be factorized using the following identities.

1. \((a – b)(a + b) = {a^2} – {b^2}\)
2. \({(a + b)^2} = {a^2} + 2ab + {b^2}\)
3. \({(a – b)^2} = {a^2} – 2ab + {b^2}\)
4. \({(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\)
5. \({(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} = {a^3} + {b^3} + 3ab(a + b)\)
6. \({(a – b)^3} = {a^3} – 3{a^2}b + 3a{b^2} – {b^3} = {a^3} – {b^3} + 3ab(a – b)\)
7. \(\left( {a – b} \right) \times \left( {{a^2} + ab + {b^2}} \right) = {a^3} – {b^3}\)
8. \(\left( {a + b} \right) \times \left( {{a^2}+ab + {b^2}} \right) = {a^3} + {b^3}\)
9. \(\left( {a + b + c} \right) \times \left( {{a^2} + {b^2} + {c^2} – ab – bc – ca} \right) = {a^3}+{b^3} + {c^3} – 3\,abc\)

In factorizing a given expression, the above-given identities can be used, whichever is required. First of all, we observe the given expression. If it is of the type of RHS of any of the above-given identities, then LHS of that identity will be the required factors.

Problems based on factorisation of trinomials which are perfect squares:

Working Rule:

Arrange the terms of the given trinomial in any of the following from whichever is suitable.

1. \({a^2} + 2ab + {b^2} = {(a + b)^2} = (a + b)(a + b)\)
2. \({a^2} – 2ab + {b^2} = {(a – b)^2} = (a – b)(a – b)\)

Problems based on factorisation of expressions which can be expressed in the form \(({a^2} – {b^2})\)

Working Rule:

If the expression is of the form \(({a^2} – {b^2})\) use the formula \({a^2} – {b^2} = (a + b)(a – b)\)

If the expression is not of the form \(({a^2} – {b^2})\) try to transform the given expression in the form \(({a^2} – {b^2})\) and then use formula \({a^2} – {b^2} = (a + b)(a – b)\)

Go on using the formula \({a^2} – {b^2} = (a + b)(a – b)\) till the given expression is completely factored.

Factorisation of quadratic equation splitting the middle terms

Problems based on factorisation of the expression of the form \({x^2} + px + q\)

If any factor is common in all terms of the given expression, take this factor out and write the remaining expression in brackets.

Find \(p\) which is the coefficient of \(x\), and \(q\), which is the constant term.

Find out factors \(a\) and \(b\) of \(q\) whose sum is \(p\).

Then write \({x^2} + px + q = {x^2} + (a + b)x + q = ({x^2} + ax) + (bx + q)\)and take out common factors in each bracket.

How to find \(a\) and \(b\):

\(p\) will be split into \(a\) and \(b\) in such a way that we have \(p = a + b\) and \(q = a \times b.\)

If the sign of \(q\) is positive, then both the factors \(a\) and \(b\) of \(q\) will have the sign same as that \(p,\,i.e.,\,if\,p\) is positive, then both \(a\) and \(b\) will be positive, but if \(p\) is negative, then both \(a\) and \(b\) will be negative.

If \(q\) is negative, then find \(\left| q \right|\) (Numerical value of \(q\)) then the numerically greater factor of \(q\) will have sign same as that of \(p\) and smaller factor will have sign opposite to that of \(p\).

Solved Examples – Methods of Factorisation

Q.1. Factorise \(6ab – 9bc\)
Ans: : The given algebraic expression is \(6ab – 9bc\)
The given expression we can write as \(2 \times 3 \times a \times b – 3 \times 3 \times b \times c = 3b(2a – 3c)\)
Hence, the factorisation of \(6ab – 9bc\) is \(3b(2a – 3c).\)

Q.2. Factorise \({a^2} + bc + ab + ac\)
Ans: By suitably rearranging the terms, we have:
\({a^2} + bc + ab + ac = {a^2} + ab + ac + bc\)
\( = a(a + b) + c(a + b)\)
\( = (a + b)(a + c)\)
Hence, the factorisation of \( = \,{a^2} + bc + ab + ac\) is \((a + b)(a + c).\)

Q.3. Factorise \(49{x^2} – 16{y^2}.\)
Ans: The given algebraic expression is \(49{x^2} – 16{y^2}.\)
The given algebraic expression can be written as \({(7x)^2} – {(4y)^2}.\)
\( = \)\((7x + 4y) – (7x – 4y)\)
Hence, the factorisation of \(49{x^2} – 16{y^2}\) is \((7x + 4y)(7x – 4y).\)

Q.4. Factorise \({x^2} + 10x + 25\)
Ans: The given algebraic expression is \({x^2} + 10x + 25\)
By suitably rewriting the terms of the given expression, we have:
\({x^2} + 10x + 25 = {x^2} + 2 \times 5 \times x + {5^2}\)
\( = {(x + 5)^2} = (x + 5)(x + 5)\)
Hence, the factorisation of \( = {x^2} + 10x + 25\) is \((x + 5)(x + 5).\)

Q.5. Factorise \({x^2} + 8x + 15\)
Ans: The given expression is \({x^2} + 8x + 15\)
Find two numbers whose sum \( = 8\) and product \( = 15\)
Clearly, the numbers are \(5\) and \(3\)
\(\therefore \,\,{x^2} + 8x + 15 = {x^2} + 5x + 3x + 15\)
\( = x(x + 5) + 3(x + 5)\)
\( = (x + 5)(x + 3)\)
Hence, the factorisation of \( = \,{x^2} + 8x + 15\) is \((x + 5)(x + 3).\)

Summary

In this article, we learned about the definition, methods of factorisation, types of methods of factorisation, notes on methods of factorisation, solved examples on methods of factorisation, frequently asked questions on methods of factorisation.

The learning outcome of this article is how to break the complex expression into a product of simple and similar objects.

Frequently Asked Questions (FAQ) – Methods of Factorisation

Let’s look at some of the commonly asked questions about methods of factorisation:

Q.1. What are the \(5\) types of factoring?
Ans: The \(5\) types of factoring are
Prime factorisation of numbers
1. Prime factorisation of numbers using factor tree method
2. Prime factorisation of numbers using repeated division method
Factorisation of algebraic expression
3. Factorisation of algebraic expressions using the method of taking common factors
4. Factorisation of algebraic expressions using by regrouping terms
5. Factorisation of quadratic expression using the method of splitting the middle term.

Q.2. Explain methods of factorisation with example?
Ans: Prime factorisation: The process of writing a given number as a prime factor product is called a prime factorisation or complete factorisation of the given number.
Example: Prime factorisation of \(54\) is \(2 \times 3 \times 3 \times 3.\)
Factorisation of algebraic expression: The process of writing a given algebraic expression as the product of two or more factors is called factorisation of algebraic expressions.
Example: Factorisation of \(4xy{z^2} = 2 \times 2 \times x \times y \times z \times z.\)

Q.3. What is meant by factoring?
Ans: Factorisation or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler things of the same kind.

Q.4. What are the \(7\) factoring techniques?
Ans: The \(7\) factoring techniques are
1. Prime factorisation of numbers
2. Factorisation of algebraic expressions when a common monomial factor occurs in each term
3. Factorisation of algebraic expressions when a binomial is a common factor
4. Factorisation by grouping the terms
5. Factorisation of binomial expressions expressible as the difference of two squares
6. Factorisation of algebraic expressions expressible as a perfect square
7. Splitting the middle terms

Q.5. What are the \(4\) methods of factoring?
Ans: There are \(4\) methods to factorize algebraic expressions
1.Taking common factors in algebraic expressions
2. Regrouping terms of algebraic expressions
3. Using standard identities
4. Splitting the middle term of the quadratic expression

NCERT Solutions For Class 8 Maths Chapter 14

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