| --- |
| 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. |
