source: ncert
topic: algebra
class_level: class_8
chapter: ch14
difficulty: intermediate
Class 8 | Ch14: Factorisation METHODS OF FACTORISATION: Method 1 - Common Factor: take out HCF of all terms. 12x^2y-9xy^2+6xyz = 3xy(4x-3y+2z). Method 2 - Grouping: group terms to find common factor. ax+ay+bx+by = a(x+y)+b(x+y) = (a+b)(x+y). a^2-b^2+a-b = (a-b)(a+b)+(a-b) = (a-b)(a+b+1). Method 3 - Using Identities: x^2+8x+16 = x^2+2(4)x+4^2 = (x+4)^2. 4x^2-9 = (2x)^2-3^2 = (2x+3)(2x-3). Method 4 - Splitting Middle Term (for quadratics ax^2+bx+c): Find two numbers p,q such that p+q=b and p×q=a×c. x^2+7x+12: p+q=7, p×q=12 → p=3,q=4. So x^2+3x+4x+12=x(x+3)+4(x+3)=(x+3)(x+4). 2x^2+7x+3: p+q=7, p×q=2×3=6 → p=6,q=1. So 2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3). DIVISION OF ALGEBRAIC EXPRESSIONS: Polynomial ÷ Monomial: divide each term separately. (6x^3-4x^2+2x) ÷ 2x = 3x^2-2x+1. Polynomial ÷ Polynomial: factorise numerator and denominator, cancel common factors. SOLVED EXAMPLES: Example 1: Factorise 12x^2y-9xy^2+6xyz = 3xy(4x-3y+2z). Example 2: x^2+7x+12 = (x+3)(x+4). Verify: (x+3)(x+4)=x^2+7x+12. Correct. Example 3: 2x^2+7x+3. Product=6, sum=7. Factors: 6 and 1. = (2x+1)(x+3). COMMON MISTAKES: Always take out the COMPLETE HCF in method 1 (common factor). When splitting middle term: product must equal a×c (coefficient of x^2 × constant). Verify factorisation by expanding back.