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Factorisation - Method of Common Factors

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Factorisation is the process of writing an algebraic expression as a product of two or more expressions. It is the reverse of the Distributive Law. Visualise this like taking a complex machine and dismantling it into its original component parts or 'factors'.

Factors of Monomials: A monomial can be broken down into its irreducible factors. For example, 5xy5xy has factors 5,x,5, x, and yy. These are called 'irreducible' because they cannot be factorised further, much like prime numbers in arithmetic.

Method of Common Factors: This involves identifying the Highest Common Factor (HCF) of all the terms in the expression. Think of this as looking at a row of boxes and picking out the exact same item that appears in every single box to place it outside.

The Distributive Property in Reverse: In multiplication, we use a(b+c)=ab+aca(b + c) = ab + ac. In factorisation, we observe the common aa in ab+acab + ac and pull it out to get a(b+c)a(b + c). This can be visualised as finding the common width 'a' for two adjacent rectangles with areas abab and acac.

Identifying Numerical HCF: For the numerical coefficients, find the largest number that divides all of them. For example, in 12x+1812x + 18, the HCF of 1212 and 1818 is 66. On a number line, this is the largest step size that lands exactly on both 1212 and 1818.

Identifying Literal (Variable) HCF: For the variables, look for the lowest power of each common variable. In x3x^{3} and x2x^{2}, the common factor is x2x^{2} because it is the largest power contained within both terms.

Irreducible Form: An expression is said to be in its factorised form when no further common factors can be extracted. Visually, this is represented as a product of brackets where the contents inside each bracket have no common factors remaining.

📐Formulae

ab+ac=a(b+c)ab + ac = a(b + c)

abac=a(bc)ab - ac = a(b - c)

ax+ay+az=a(x+y+z)ax + ay + az = a(x + y + z)

HCF of terms×(Quotient of Term 1+Quotient of Term 2+)\text{HCF of terms} \times (\text{Quotient of Term 1} + \text{Quotient of Term 2} + \dots)

💡Examples

Problem 1:

Factorise the expression: 12x2y+15xy212x^{2}y + 15xy^{2}

Solution:

Step 1: Find the factors of each term. 12x2y=2×2×3×x×x×y12x^{2}y = 2 \times 2 \times 3 \times x \times x \times y 15xy2=3×5×x×y×y15xy^{2} = 3 \times 5 \times x \times y \times y Step 2: Identify the common factors. The common factors are 3,x,3, x, and yy. Step 3: Calculate the HCF. HCF=3×x×y=3xy\text{HCF} = 3 \times x \times y = 3xy Step 4: Write the expression as product of HCF and the remaining terms. 12x2y+15xy2=3xy(4x+5y)12x^{2}y + 15xy^{2} = 3xy(4x + 5y)

Explanation:

We first break down both terms into their prime and variable factors. By circling the components that appear in both lists (3,x,y3, x, y), we determine the HCF. Dividing each original term by 3xy3xy gives us the terms left inside the bracket.

Problem 2:

Factorise 10a215b2+20c210a^{2} - 15b^{2} + 20c^{2}

Solution:

Step 1: Find the numerical HCF of the coefficients 10,15,10, 15, and 2020. Factors of 10:2,510: 2, 5 Factors of 15:3,515: 3, 5 Factors of 20:2,2,520: 2, 2, 5 The common numerical factor is 55. Step 2: Check for literal (variable) common factors. There are no variables (a,b,ca, b, c) common to all three terms. Step 3: Factor out the 55. 10a215b2+20c2=5(2a23b2+4c2)10a^{2} - 15b^{2} + 20c^{2} = 5(2a^{2} - 3b^{2} + 4c^{2})

Explanation:

In this problem, there are no common variables across all terms. We only find the HCF of the numbers 10,15,10, 15, and 2020, which is 55, and divide each term by 55 to find the expression inside the parentheses.