ICSE Class 8 Maths Algebra Chapter 05 Factorization

Factorization

If a given algebraic expression can be expressed as a product of two or more expressions then those expressions are called factors of the given expression.

Examples

(i) \(6 + 2x = 2(3 + x)\), so 2 and \(3 + x\) are factors of \(6 + 2x\).

(ii) \(a^2 - b^2 = (a + b)(a - b)\), so \(a + b\) and \(a - b\) are factors of \(a^2 - b^2\).

(iii) \((x + 1)(x + 2) = x^2 + 3x + 2\), so \(x + 1\) and \(x + 2\) are factors of \(x^2 + 3x + 2\).

Methods Of Factorization

To resolve an expression into its factors is called factorization. The method we use to factorize an expression depends on the type of expression. In this chapter, we will take a look at a few methods of factorization used commonly.

Taking Out Common Factors

Follow these steps when all the terms of a polynomial have common factors.

1. Find the highest common factor (HCF) of the terms by (i) finding the HCF of the numerical coefficients of the terms, (ii) finding the highest power of each variable common to the terms, and (iii) multiplying all the common factors.

2. Divide each term of the polynomial by this common factor.

3. Write the quotient within brackets and the common factor outside the brackets.

Solved Examples

Example 1

Factorize \(4xy - 8y^2\).

Solution

\(4xy = 4 \times x \times y\), \(8y^2 = 4 \times 2 \times y \times y\).

The HCF of the numerical coefficients of the terms = 4.

The highest power of \(y\) common to all the terms = \(y\).

Therefore, the HCF of the terms = \(4y\).

Dividing the expression by the common factor, \(\frac{4xy}{4y} - \frac{8y^2}{4y} = x - 2y\).

Therefore, \(4xy - 8y^2 = 4y(x - 2y)\).

Example 2

Factorize \(3a^2b^2 - 9ab^3 + 15ab^4\).

Solution

\(3a^2b^2 = 3 \times a \times a \times b \times b\), \(9ab^3 = 3 \times 3 \times a \times b \times b \times b\), \(15ab^4 = 3 \times 5 \times a \times b \times b \times b \times b\)

The HCF of the numerical coefficients = 3.

The highest common power of \(a\) = \(a\).

The highest common power of \(b\) = \(b^2\).

Therefore, the HCF of the terms = \(3ab^2\).

\(3a^2b^2 - 9ab^3 + 15ab^4 = 3ab^2\left(\frac{3a^2b^2}{3ab^2} - \frac{9ab^3}{3ab^2} + \frac{15ab^4}{3ab^2}\right) = 3ab^2(a - 3b + 5b^2)\).

Example 3

Factorize \(2(x + y) + 3(x + y)^2\).

Solution

The HCF of the numerical coefficients = 1.

The highest power of the binomial \((x + y)\) common to both the terms = \(x + y\).

Therefore, the HCF of the terms = \(1 \times (x + y) = x + y\).

\(2(x + y) + 3(x + y)^2 = (x + y)\left(\frac{2(x + y)}{x + y} + \frac{3(x + y)^2}{x + y}\right)\)

\(= (x + y)\{2 + 3(x + y)\} = (x + y)(3x + 3y + 2)\).

Example 4

Factorize \(5(3a + 2b)^2 - 20(3a + 2b)^4 + 60(3a + 2b)^5\).

Solution

The HCF of 5, 20 and 60 = 5.

The highest power of the binomial \((3a + 2b)\) common to all the terms = \((3a + 2b)^2\).

Therefore, the HCF of the terms = \(5 \times (3a + 2b)^2 = 5(3a + 2b)^2\).

\(5(3a + 2b)^2 - 20(3a + 2b)^4 + 60(3a + 2b)^5\)

\(= 5(3a + 2b)^2\left\{\frac{5(3a + 2b)^2 - 20(3a + 2b)^4 + 60(3a + 2b)^5}{5(3a + 2b)^2}\right\}\)

\(= 5(3a + 2b)^2\{1 - 4(3a + 2b)^2 + 12(3a + 2b)^3\}\).

Teacher's Note

Factorization helps us simplify complex expressions, much like breaking down a large task into smaller, manageable steps makes it easier to complete.

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