Factorisation for CBSE Class 8 Maths

What is Factorisation?

Factorisation is the process of writing an algebraic expression as a product of two or more algebraic expressions, called its factors. These factors can be numbers, variables, or other algebraic expressions. For example, consider the expression 5xy + 10x. We can see that both terms have 5 and x in common. By taking 5x as a common factor, we can write the expression as 5x(y + 2). Here, 5, x, and (y + 2) are the factors of 5xy + 10x. The goal is to find factors that are 'irreducible', meaning they cannot be factorised any further. This is very similar to finding the prime factors of a number. Just as the prime factors of 12 are 2, 2, and 3, the irreducible factors of 5xy + 10x are 5, x, and (y+2). Mastering this skill allows us to simplify problems and see the underlying structure of equations.

Method 1: Finding Common Factors

1. Step 1: Decompose each term — Break down every term in the expression into its irreducible factors (prime numbers and individual variables). For example, to factorise 8x²y + 12xy², we write each term's factors: 8x²y = 2 × 2 × 2 × x × x × y 12xy² = 2 × 2 × 3 × x × y × y
2. Step 2: Identify the common factors — Look for the factors that appear in the decomposition of all the terms. In our example, the common factors are 2, 2, x, and y. The product of these common factors is the Highest Common Factor (HCF), which is 2 × 2 × x × y = 4xy.
3. Step 3: Factor out the HCF — Use the distributive law in reverse (ab + ac = a(b+c)) to pull out the HCF. Divide each original term by the HCF to find what remains inside the bracket. 8x²y + 12xy² = 4xy (8x²y / 4xy) + 4xy (12xy² / 4xy) = 4xy(2x) + 4xy(3y) = 4xy(2x + 3y) The factors are 4xy and (2x + 3y).

Method 2: Factorisation by Regrouping Terms

• Sometimes, there is no single factor common to all terms. In such cases, we can try grouping the terms to find common factors within smaller groups. Problem: Factorise the expression ax + bx + ay + by. Step 1: Observe the terms. There is no factor common to all four terms. Step 2: Group the terms. Let's group the first two terms and the last two terms together: (ax + bx) + (ay + by). Step 3: Factor out the common factor from each group. In the first group (ax + bx), the common factor is x. So, we get x(a + b). In the second group (ay + by), the common factor is y. So, we get y(a + b). Step 4: Combine the results. Our expression now looks like this: x(a + b) + y(a + b). Step 5: Factor out the common binomial factor. Notice that (a + b) is now a common factor to the entire expression. We can factor it out. (a + b)(x + y). Final Answer: The factors of ax + bx + ay + by are (a + b) and (x + y).

Method 3: Factorisation using Identities

• Identity I: a² + 2ab + b² = (a + b)². If an expression is a perfect square trinomial with a positive middle term, its factors are (a + b) and (a + b). Example: x² + 6x + 9 = (x + 3)².
• Identity II: a² – 2ab + b² = (a – b)². If an expression is a perfect square trinomial with a negative middle term, its factors are (a – b) and (a – b). Example: y² – 10y + 25 = (y – 5)².
• Identity III: a² – b² = (a + b)(a – b). This is the 'difference of squares'. If you have one perfect square subtracted from another, it factors easily. Example: 4x² – 9 = (2x)² – 3² = (2x + 3)(2x – 3).
• Identity IV: x² + (a + b)x + ab = (x + a)(x + b). For trinomials, look for two numbers (a and b) that multiply to give the constant term (ab) and add to give the coefficient of x (a+b). Example: For p² + 6p + 8, we need two numbers that add to 6 and multiply to 8. The numbers are 4 and 2. So, the factors are (p + 4)(p + 2).

Practice Questions with Solutions

• Q: Factorise the expression: 12a²b + 15ab² A: Step 1: Find the prime factors of each term. 12a²b = 2 × 2 × 3 × a × a × b 15ab² = 3 × 5 × a × b × b Step 2: Identify the highest common factors (HCF). The common factors are 3, a, and b. The HCF is 3ab. Step 3: Factor out the HCF from the expression. 12a²b + 15ab² = 3ab(4a) + 3ab(5b) = 3ab(4a + 5b) Final answer: 3ab(4a + 5b)
• Q: Factorise the expression: 6xy - 4y + 6 - 9x A: Step 1: The terms do not have a common factor, so we try regrouping. Let's rearrange for convenience: 6xy - 9x - 4y + 6. Step 2: Group the terms: (6xy - 9x) + (-4y + 6). Step 3: Factor out the HCF from each group. From (6xy - 9x), the HCF is 3x, giving 3x(2y - 3). From (-4y + 6), the HCF is -2, giving -2(2y - 3). Step 4: The expression becomes 3x(2y - 3) - 2(2y - 3). The common binomial factor is (2y - 3). Step 5: Factor out the common binomial: (2y - 3)(3x - 2). Final answer: (2y - 3)(3x - 2)
• Q: Factorise the expression: 49p² - 36 A: Step 1: Observe the expression. It has two terms, both are perfect squares, and they are separated by a minus sign. This fits the identity a² - b² = (a + b)(a - b). Step 2: Write each term as a square. 49p² = (7p)² and 36 = 6². Step 3: Apply the identity. Here, a = 7p and b = 6. (7p)² - 6² = (7p + 6)(7p - 6). Final answer: (7p + 6)(7p - 6)
• Q: Factorise the expression: m² + 7m + 10 A: Step 1: This is a trinomial of the form x² + (a+b)x + ab. We need to find two numbers, a and b, such that their sum (a+b) is 7 and their product (ab) is 10. Step 2: Think of pairs of numbers that multiply to 10: (1, 10), (2, 5). Now check their sums: 1+10=11, 2+5=7. The correct pair is 2 and 5. Step 3: Apply the identity (x+a)(x+b). Here, x=m, a=2, and b=5. m² + 7m + 10 = (m + 2)(m + 5). Final answer: (m + 2)(m + 5)

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