ICSE Class 8 Maths Algebra Chapter 17 Factorisaion PDF Download

Factorisation

Factorisation Of Algebraic Expressions

Factorisation of an algebraic expression means finding two or more expressions which when multiplied give the original expression as the product.

Thus, factorisation is the opposite of finding the product, as illustrated by the following examples.

In class VII we factorised expressions by taking out the HCF as the common factor, by the grouping method and by expressing the product as a difference of squares, as recapitulated in the following examples.

Method I: Factorisation By Taking Out The Common Factor

Example 1: Factorise \(2x^4y^2 - 8x^3y^3 + 6x^3y^3 - 4x^3y^2\)

Find the HCF of all the terms in the expression \(2x^4y^2 = 2 \times x \times x \times x \times x \times y \times y\) \(8x^4y^3 = 2 \times 2 \times 2 \times x \times x \times x \times x \times y \times y \times y\) \(6x^3y^3 = 2 \times 3 \times x \times x \times x \times x \times y \times y \times y\) \(4x^3y^2 = 2 \times 2 \times x \times x \times x \times x \times y \times y\)

Thus, HCF of all the terms in the polynomial = \(2 \times x \times x \times x \times y \times y = 2x^3y^2\)

Dividing the polynomial by the HCF, we get \[\frac{2x^4y^2}{2x^3y^2} - \frac{8x^4y^3}{2x^3y^2} + \frac{6x^3y^3}{2x^3y^2} - \frac{4x^3y^2}{2x^3y^2}\] \[= x - 4y + 3y - 2\]

Thus, \(2x^4y^2 - 8x^4y^3 + 6x^3y^3 - 4x^3y^2 = 2x^3y^2(x - 4y + 3y - 2)\)

Teacher's Note

Factorisation helps us break down complex algebraic expressions into simpler components, much like dividing a large project into manageable tasks makes work easier.

Method II: Factorisation By The Grouping Method

Example 2: Factorise \(18xy - 9y + 30x - 15\)

\(= 3(6xy - 3y + 10x - 5)\) (taking 3 as a common factor)

\(= 3(6xy + 10x - 3y - 5)\) (grouping terms with common factor)

\(= 3\{2x(3y + 5) - 1(3y + 5)\}\) \(= 3(3y + 5)(2x - 1)\) [taking (3y + 5) as common factor]

Example 3: Factorise \(2x^5 + 9x^4 - 3x^3 + 3x - 6x^4 - 9\)

\(= 2x^5 - 6x^4 - 3x^3 + 9x^2 + 3x - 9\) (rearranging terms in decreasing powers of variable)

\(= 2x^4(x - 3) - 3x^2(x - 3) + 3(x - 3)\) (taking common factors from pairs)

\(= (x - 3)(2x^4 - 3x^2 + 3)\) [taking (x - 3) as common factor]

Example 4: Factorise \(2x^3 + 6y + 3x^2 + 4x^2y + 8x + 12\)

The terms can be rearranged in decreasing powers of x in more than one way.

\(2x^3 + 3x^2 + 4x^2y + 8x + 6y + 12\)

Taking common factors from pairs we get

\(x^2(2x + 3) + 4x(xy + 2) + 6(xy + 2)\)

But this grouping does not help as the factors left are different. The terms have to be rearranged in pairs in such a way that after dividing each pair by the common factors, the same factor is left.

Rearranging as:

\(2x^3 + 3x^2 + 4x^2y + 6xy + 8x + 12\) \(= x^2(2x + 3) + 2xy(2x + 3) + 4(2x + 3)\) \(= (2x + 3)(x^2 + 2xy + 4)\)

Observe that the factorisation of the first pair as x²(2x + 3) gave an indication on how the other terms would be grouped such that after each group was divided by a common factor, (2x + 3) would be left as the other factor.

Teacher's Note

The grouping method in factorisation teaches us how rearranging and organizing information strategically can reveal hidden patterns and simplify complex problems.

Factorising The Difference Of Two Squares

As \((a + b)(a - b) = a^2 - b^2\), \(a^2 - b^2 = (a + b)(a - b)\)

Example 5: Factorise \(\frac{8x^2b}{27z^2} - \frac{18y^2b}{12}\)

\(\frac{8x^2b}{27z^2} - \frac{18y^2b}{12} = \frac{2b}{3}\left(\frac{4x^2}{9z^2} - \frac{9y^2}{4}\right)\) (taking \(\frac{2b}{3}\) as common factor)

As \(\frac{4x^2}{9z^2} = \left(\frac{2x}{3z}\right)^2\) and \(\frac{9y^2}{4} = \left(\frac{3y}{2}\right)^2\),

\(\frac{2b}{3}\left(\frac{4x^2}{9z^2} - \frac{9y^2}{4}\right) = \frac{2b}{3}\left(\frac{2x}{3z} + \frac{3y}{2}\right)\left(\frac{2x}{3z} - \frac{3y}{2}\right)\)

Example 6: Factorise \(81x^4 - 16\)

\(81x^4 - 16 = (9x^2)^2 - 4^2\) \(= (9x^2 + 4)(9x^2 - 4)\) \(= (9x^2 + 4)\{(3x)^2 - 2^2\}\) \(= (9x^2 + 4)(3x + 2)(3x - 2)\)

Try this!

1. Factorise \(7x + 8xy\)

2. Factorise \(144x^4 - 12t\)

Teacher's Note

The difference of squares formula is like recognizing that subtracting two perfect squares reveals factors - a pattern we can spot in many real-world measurements and calculations.

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