ICSE Class 7 Maths Chapter 19 Fractorisation

Chapter 19: Factorisation

Basic Concept

Since, the product of 5, x and y is 5xy, 5xy is divisible by 5, x and y and so we say: the factors of term 5xy are 5, x and y.

Similarly, the product of x² and (x + 3y)

= x²(x + 3y) = x³ + 3x²y

the factors of x³ + 3x²y are x² and x + 3y.

And, we write: x³ + 3x²y = x²(x + 3y)

Factors And Factorisation

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

The process of finding the factors of a given expression is called factorisation.

For example:

Different Types Of Factorizations

Type 1: Taking Out The Common Factor

Steps:

1. Find, by inspection, the greatest monomial by which each term of the given expression can be divided completely.

2. Divide each term by this monomial and enclose the quotient within a bracket, keeping the common monomial outside the bracket.

Example 1

Factorise: 4xy - 8y²

Solution

Step 1: The given expression 4xy - 8y² has two terms 4xy and 8y². The greatest monomial by which these terms can completely be divided is 4y.

Step 2: 4xy - 8y² = 4y\(\left(\frac{4xy}{4y} - \frac{8y^2}{4y}\right)\)

[Dividing each term by 4y and keeping 4y outside the bracket]

= 4y(x - 2y)

Alternative method:

Step 1: Express each term of the given expression as a product of all its factors.

Step 2: From the factors, obtained in step 1, find out the highest common factor.

Step 3: Divide each term of the given expression by the highest common factor, obtained in step 2, and enclose the quotient within a bracket, keeping the highest common factor, obtained in step 2, outside the bracket.

Since,

4xy = 2 × 2 × x × y

and

8y² = 2 × 2 × 2 × y × y

Highest common factor = 2 × 2 × y = 4y

.\(\therefore\) 4xy - 8y² = 4y\(\left(\frac{4xy}{4y} - \frac{8y^2}{4y}\right)\)

= 4y(x - 2y)

Example 2

Factorise: 6a² + 9ab + 12a

Solution

The greatest monomial by which all the terms of the given expression can be divided completely is 3a.

.\(\therefore\) 6a² + 9ab + 12a = 3a\(\left(\frac{6a^2}{3a} + \frac{9ab}{3a} + \frac{12a}{3a}\right)\)

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

Since,

6a² = 2 × 3 × a × a

9ab = 3 × 3 × a × b and

12a = 2 × 2 × 3 × a

H.C.F. = 3 × a = 3a

Example 3

Factorise: (i) 4x² + 5xy - 6xy² (ii) -3a² - 6ab + 12ab²

Solution

(i) 4x² + 5xy - 6xy² = x\(\left(\frac{4x^2}{x} + \frac{5xy}{x} - \frac{6xy^2}{x}\right)\)

= x(4x + 5y - 6y²)

(ii) -3a² - 6ab + 12ab² = -3a\(\left(\frac{-3a^2}{-3a} - \frac{6ab}{-3a} + \frac{12ab^2}{-3a}\right)\)

= -3a(a + 2b - 4b²)

Teacher's Note

Finding common factors helps us simplify expressions, much like finding common denominators when adding fractions, which you encounter in calculating costs and measurements in daily situations.

Type 2: Grouping

Consider the expression: ax + bx + ay + by.

This expression has no common factor on the whole, but on observing carefully, we find that the first two terms have x as a common factor and the last two terms have y as a common factor.

Such an expression can be resolved into factors by adopting the following steps:

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

2. Factorise each group.

3. Take out the factor which is common to each group.

Thus, ax + bx + ay + by = (ax + bx) + (ay + by) [Step 1]

= x(a + b) + y(a + b) [Step 2]

= (a + b)(x + y) [Step 3]

Factorisation by grouping is possible only if the given expression has an even number of terms and the minimum number of terms in it is 4.

Example 4

Factorise: (i) 3x³ - 6x² + ax - 2a (ii) 5ph - 10qk + 2rph - 4qrk

Solution

(i) 3x³ - 6x² + ax - 2a = (3x³ - 6x²) + (ax - 2a) [Step 1]

= 3x²(x - 2) + a(x - 2) [Step 2]

= (x - 2)(3x² + a) [Step 3]

(ii) 5ph - 10qk + 2rph - 4qrk = (5ph - 10qk) + (2rph - 4qrk)

= 5(ph - 2qk) + 2r(ph - 2qk)

= (ph - 2qk)(5 + 2r)

OR,

5ph - 10qk + 2rph - 4qrk = 5ph + 2rph - 10qk - 4qrk [On re-arranging]

= (5ph + 2rph) - (10qk + 4qrk) [Step 1]

= ph(5 + 2r) - 2qk(5 + 2r) [Step 2]

= (5 + 2r)(ph - 2qk) [Step 3]

Teacher's Note

Grouping strategies help organize complex problems, similar to how you categorize items when organizing your study materials or arranging items in a store.

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