ICSE Class 9 Maths Chapter 05 Factorisation

5 Factorisation

5.1 Introduction

When a polynomial (an algebraic expression) is expressed as the product of two or more expressions, each of these expressions is called a factor of the polynomial.

The polynomial \(x^2 + 5x + 6\) can be expressed as the product of the expressions \((x + 3)\) and \((x + 2)\).

That is \(x^2 + 5x + 6 = (x + 3)(x + 2) \Rightarrow (x + 3)\) and \((x + 2)\) are factors of \(x^2 + 5x + 6\).

The process of writing an expression in the form of terms or brackets multiplied together, is called factorisation. Each term and each bracket is called a factor of the expression.

e.g. (i) \(5x^2 + 15 = 5(x^2 + 3)\)

\(\Rightarrow\) 5 and \(x^2 + 3\) are factors of \(5x^2 + 15\).

(ii) \(ax^2 + 5ax + 6a = a(x + 3)(x + 2)\)

\(\Rightarrow\) a, \((x + 3)\) and \((x + 2)\) are factors of \(ax^2 + 5ax + 6a\).

Factorisation is the reverse of multiplication.

Teacher's Note

Factorisation is like taking apart a LEGO structure into its original building blocks - understanding how complex expressions break down into simpler components helps solve real-world problems in engineering and science.

5.2 Methods Of Factorisation

Type 1: Taking Out The Common Factors

When each term of a given expression contains a common factor, divide each term by this factor and enclose the quotient within brackets, keeping the common factor outside the bracket.

Procedure:

Find the H.C.F. of all the terms of the given expression.

For expression \(6a^2 - 3ax\), its terms are \(6a^2\) and \(-3ax\). And, the H.C.F. of these terms is \(3a\).

Therefore, \(6a^2 - 3ax = 3a\left(\frac{6a^2}{3a} - \frac{3ax}{3a}\right) = 3a(2a - x)\).

Factorise:

(i) \(8ab^2 + 12a^2b\)

(ii) \(4(x + y)^2 - 3(x + y)\)

(iii) \(x(a - 5) + y(5 - a)\)

Solution:

(i) \(8ab^2 + 12a^2b = 4ab\left(\frac{8ab^2}{4ab} + \frac{12a^2b}{4ab}\right)\) [H.C.F. of \(8ab^2\) and \(12a^2b\) is \(4ab\)]

\(= 4ab(2b + 3a)\)

Direct method: It can easily be seen that \(4ab\) is the largest expression which divides both the terms \(8ab^2\) and \(12a^2b\) of the given expression \(8ab^2 + 12a^2b\) completely.

\(\therefore\) \(8ab^2 + 12a^2b = 4ab(2b + 3a)\)

(ii) \(4(x + y)^2 - 3(x + y) = (x + y)[4(x + y) - 3]\)

\(= (x + y)(4x + 4y - 3)\)

(iii) \(x(a - 5) + y(5 - a) = x(a - 5) - y(a - 5)\)

\(= (a - 5)(x - y)\)

Teacher's Note

Finding common factors is like identifying shared characteristics among people - in algebra, we extract what's common to simplify complex expressions, much like grouping similar items in everyday organization.

Type 2: Grouping

An expression of an even number of terms, may be resolved into factors, if the terms are arranged in groups such that each group has a common factor.

Procedure:

1. Group the terms of the given expression in such a way that each group has a common factor.

2. Factorise each group formed.

3. From each group, obtained in step 2, take out the common factor.

Factorise:

(i) \(ab + bc + ax + cx\)

(ii) \(ab^2 - (a - 1)b - 1\)

Solution:

(i) \(ab + bc + ax + cx = (ab + bc) + (ax + cx)\) [Forming groups]

\(= b(a + c) + x(a + c)\) [Taking out common factors from each group]

\(= (a + c)(b + x)\) [Taking \((a + c)\) common]

(ii) \(ab^2 - (a - 1)b - 1 = ab^2 - ab + b - 1\)

\(= ab(b - 1) + 1(b - 1)\)

\(= (b - 1)(ab + 1)\)

Factorise: \(a^2 + \frac{1}{a^2} + 2 - 5a - \frac{5}{a}\)

Solution:

\(a^2 + \frac{1}{a^2} + 2 - 5a - \frac{5}{a} = \left(a^2 + \frac{1}{a^2} + 2\right) - 5\left(a + \frac{1}{a}\right)\)

\(= \left(a + \frac{1}{a}\right)^2 - 5\left(a + \frac{1}{a}\right) = \left(a + \frac{1}{a}\right)\left(a + \frac{1}{a} - 5\right)\)

Teacher's Note

Grouping terms in algebra is like sorting laundry by color before washing - organizing similar items together makes the overall process much simpler and more efficient.

Exercise 5 (A)

Factorise by taking out the common factors:

1. \(3a^2 - 9ab\)

2. \(2(x + y)^3 - 6(x + y)\)

3. \(x^3(2x - 3y) - x^2(2x - 3y)^2\)

4. \(2(2x - 5y)(3x + 4y) - 6(2x - 5y)(x - y)\)

Factorise by grouping method:

5. \(a^3 + a - 3a^2 - 3\)

6. \(16(a + b)^2 - 4a - 4b\)

7. \(a^4 - 2a^3 - 4a + 8\)

8. \(ab - 2b + a^2 - 2a\)

9. \(ab(x^2 + 1) + x(a^2 + b^2)\)

10. \(a^2 + b - ab - a\)

11. \((ax + by)^2 + (bx - ay)^2\)

12. \(a^2x^2 + (ax^2 + 1)x + a\)

13. \((2a - b)^2 - 10a + 5b\)

14. \(a(a - 4) - a + 4\)

15. \(y^2 - (a + b)y + ab\)

16. \(a^2 + \frac{1}{a^2} - 2 - 3a + \frac{3}{a}\)

17. \(x^2 + y^2 + x + y + 2xy\)

18. \(a^2 + 4b^2 - 3a + 6b - 4ab\)

19. \(m(x - 3y)^2 + n(3y - x) + 5x - 15y\)

20. \(x(6x - 5y) - 4(6x - 5y)^2\)

Type 3: Trinomial Of The Form ax² - bx + c (By Splitting The Middle Term)

When a trinomial is of the form \(ax^2 + bx + c\) (or \(a + bx + cx^2\)), split b (the coefficient of x in the middle term) into two parts such that the sum of these two parts is equal to b and the product of these two parts is equal to the product of a and c. Then factorize by the grouping method.

Factorise:

(i) \(x^2 + 5x + 6\)

(ii) \(x^2 - 5x + 6\)

(iii) \(x^2 - 5x - 6\)

(iv) \(x^2 + 5x - 6\)

Solution:

(i) \(x^2 + 5x + 6 = x^2 + 3x + 2x + 6\) Since, \(3 + 2 = 5\)

\(= x(x + 3) + 2(x + 3)\) and, \(3 \times 2 = 6\)

\(= (x + 3)(x + 2)\)

(ii) \(x^2 - 5x + 6 = x^2 - 3x - 2x + 6\) Since, \((-3) + (-2) = -5\)

\(= x(x - 3) - 2(x - 3)\) and, \((-3) \times (-2) = +6\)

\(= (x - 3)(x - 2)\)

(iii) \(x^2 - 5x - 6 = x^2 - 6x + x - 6\) Since, \(-6 + 1 = -5\)

\(= x(x - 6) + 1(x - 6)\) and, \((-6) \times 1 = -6\)

\(= (x - 6)(x + 1)\)

(iv) \(x^2 + 5x - 6 = x^2 + 6x - x - 6\) Since, \(6 - 1 = 5\)

\(= x(x + 6) - 1(x + 6)\) and, \(6 \times (-1) = -6\)

\(= (x + 6)(x - 1)\)

Factorise:

(i) \(2x^2 - 7x + 6\)

(ii) \(3x^2 - 11x - 4\)

(iii) \(6 + 11x + 3x^2\)

(iv) \(7 - 12x - 4x^2\)

Solution:

(i) \(2x^2 - 7x + 6 = 2x^2 - 4x - 3x + 6\) Since, \(-4 -3 = -7\) and, \((-4) \times (-3) = 12\)

\(= 2x(x - 2) - 3(x - 2) = (x - 2)(2x - 3)\)

(ii) \(3x^2 - 11x - 4 = 3x^2 - 12x + x - 4\)

\(= 3x(x - 4) + 1(x - 4) = (x - 4)(3x + 1)\)

Teacher's Note

Splitting the middle term is like breaking down a complex task into simpler steps - in algebra, we decompose expressions to find their hidden structure, much like analyzing how complex machinery works by examining its parts.

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