ICSE Class 8 Maths Chapter 14 Factorisations

Chapter 14: Factorisation

You know that the product of \(5x^2\) and \(2x - 3y = 5x^2(2x - 3y) = 10x^3 - 15x^2y\). We say that \(5x^2\) and \(2x - 3y\) are factors of \(10x^3 - 15x^2y\). We write it as

\[10x^3 - 15x^2y = 5x^2(2x - 3y).\]

Similarly, the product of \(3x + 7\) and \(3x - 7 = (3x + 7)(3x - 7) = 9x^2 - 49\); we say that \(3x + 7\) and \(3x - 7\) are factors of \(9x^2 - 49\). We write it as

\[9x^2 - 49 = (3x + 7)(3x - 7).\]

Thus, when an algebraic expression can be written as the product of two or more expressions, then each of these expressions is called a factor of the given expression.

To find factors of a given expression means to obtain two or more expressions whose product is the given expression.

The process of finding two or more expressions whose product is the given expression is called factorisation.

Thus, factorisation is the reverse process of multiplication.

For Example:

In the previous class, you have already learnt the factorisation of polynomials by the following methods:

Taking out common factors

Grouping

Difference of two squares by using \(a^2 - b^2 = (a + b)(a - b)\)

In this chapter, we shall review the above methods and solve some tougher problems. We shall also find factors of trinomials of the type

\((i) x^2 + px + q\), where \(p, q \in \mathbb{N}\)

\((ii) ax^2 + bx + c\), where \(a, b, c \in \mathbb{N}\).

Before taking up factorisation, we would like to introduce the concept of H.C.F.

H.C.F. of two or more polynomials (with integral coefficients) is the largest common factor of the given polynomials.

H.C.F. of two or more monomials = (H.C.F. of their numerical coefficients) - (H.C.F. of their literal coefficients)

H.C.F. of literal coefficients = product of each common literal raised to the lowest power

Remark

Here it is understood that the numerical coefficients of the monomials (under consideration) are integers and the powers of the literals involved in the monomials are positive integers.

For Example:

\((i)\) H.C.F. of \(6x^2y^2\) and \(8xy^3\):

H.C.F. of numerical coefficients = H.C.F. of 6 and 8 = 2

H.C.F. of literal coefficients = H.C.F. of \(x^2y^2\) and \(xy^3\) = product of each common literal raised to the lowest power = \(xy^2\)

Therefore, H.C.F. of \(6x^2y^2\) and \(8xy^3 = 2 \times xy^2 = 2xy^2\).

\((ii)\) H.C.F. of \(15a^3b^2c^3\), \(12a^2bc^4\) and \(18a^5b^3c^2\):

H.C.F. of numerical coefficients = H.C.F. of 15, 12 and 18 = 3

H.C.F. of literal coefficients = H.C.F. of \(a^3b^2c^3\), \(a^2bc^4\) and \(a^5b^3c^2\) = product of each common literal raised to the lowest power = \(a^2bc^2\)

Therefore, H.C.F. of the given monomials = \(3 \times a^2bc^2 = 3a^2bc^2\).

Factorising By Taking Out Common Factors

If the different terms/expressions of the given polynomial have common factors, then the given polynomial can be factorised by the following procedure:

\((i)\) Find the H.C.F. of all the terms/expressions of the given polynomial.

\((ii)\) Divide each term/expression of the given polynomial by H.C.F. Enclose the quotient within the brackets and keep the common factor outside the bracket.

Example 1.

Factorise the following polynomials:

\((i) 24x^3 - 32x^2\) \((ii) 15ab^2 - 21a^2b\) \((iii) 14x^2y^2 - 10x^2y + 8xy^2\).

Solution.

\((i)\) H.C.F. of \(24x^3\) and \(32x^2\) is \(8x^2\)

\(24x^3 - 32x^2 = 8x^2(3x - 4)\).

\((ii)\) H.C.F. of \(15ab^2\) and \(21a^2b\) is \(3ab\)

Therefore, \(15ab^2 - 21a^2b = 3ab(5b - 7a)\).

\((iii)\) H.C.F. of \(14x^2y^2\), \(10x^2y\) and \(8xy^2\) is \(2xy\)

Therefore, \(14x^2y^2 - 10x^2y + 8xy^2 = 2xy(7xy - 5x + 4y)\).

Example 2.

Factorise the following:

\((i) 3x(y + 2z) + 5a(y + 2z)\)

\((ii) 10(p - 2q)^3 + 6(p - 2q)^2 - 20(p - 2q)\).

Solution.

\((i)\) H.C.F. of the expressions \(3x(y + 2z)\) and \(5a(y + 2z)\) is \(y + 2z\)

Therefore, \(3x(y + 2z) + 5a(y + 2z) = (y + 2z)(3x + 5a)\)

\((ii)\) H.C.F. of the expressions \(10(p - 2q)^3\), \(6(p - 2q)^2\) and \(20(p - 2q)\) is \(2(p - 2q)\)

Therefore, \(10(p - 2q)^3 + 6(p - 2q)^2 - 20(p - 2q) = 2(p - 2q)[5(p - 2q)^2 + 3(p - 2q) - 10]\).

Teacher's Note

Factorisation helps us simplify complex algebraic expressions, similar to how we organize items in a grocery store by grouping them into categories for easier management.

Exercise 14.1

Factorise the following (1 to 9) polynomials:

1. \((i) 8xy^3 + 12x^3y^2\) \((ii) 15ax^3 - 9ax^2\)

2. \((i) 21py^2 - 56 py\) \((ii) 4x^3 - 6x^2\)

3. \((i) 2nv^2 - 4nv\) \((ii) 18m + 16n\)

4. \((i) 25abc^3 - 15a^2b^2c\) \((ii) 28p^2q^2r - 42pq^2r^2\)

5. \((i) 8x^3 - 6x^2 + 10x\) \((ii) 14mn + 22m - 62p\)

6. \((i) 18p^2q^2 - 24pq^2 + 30p^2q\) \((ii) 27a^4b^3 - 18a^2b^3 + 75a^3b^2\)

7. \((i) 15a(2p - 3q) - 10b(2p - 3q)\) \((ii) 3a(x^2 + y^2) + 6b(x^2 + y^2)\)

8. \((i) 6(x + 2y)^3 + 8(x + 2y)^2\) \((ii) 14(a - 3b)^3 - 21p(a - 3b)\)

9. \(10a(2p + q)^3 - 15b(2p + q)^2 + 35(2p + q)\)

Factorising By Grouping Of Terms

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

\((i)\) Arrange the terms of the given polynomial in groups in such a way that each group has a common factor.

\((ii)\) Factorise each group.

\((iii)\) 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 1.

Factorise the following polynomials:

\((i) ax - ay + bx - by\) \((ii) 4x^2 - 10xy - 6xz + 15yz\).

Solution.

\((i) ax - ay + bx - by = (ax - ay) + (bx - by) = a(x - y) + b(x - y) = (x - y)(a + b)\).

\((ii) 4x^2 - 10xy - 6xz + 15yz = (4x^2 - 10xy) - (6xz - 15yz) = 2x(2x - 5y) - 3z(2x - 5y) = (2x - 5y)(2x - 3z)\).

Example 2.

Factorise the following polynomials:

\((i) x^3 + 2x^2 + x + 2\) \((ii) 1 + p + pq + p^2q\).

Solution.

\((i) x^3 + 2x^2 + x + 2 = (x^3 + 2x^2) + (x + 2) = x^2(x + 2) + 1(x + 2) = (x + 2)(x^2 + 1)\).

\((ii) 1 + p + pq + p^2q = (1 + p) + (pq + p^2q) = 1(1 + p) + pq(1 + p) = (1 + p)(1 + pq)\).

Teacher's Note

Grouping terms in factorisation is like organizing a messy room by creating sections - once we group related items together, the overall structure becomes much clearer.

This is a preview of the first 3 pages. To get the complete book, click below.