Polynomials and Factorisation — Telangana (SSC) Class 9 Mathematics Solutions (Free)

Step 1: Set up long division Step 2: x³ ÷ x = x² Step 3: x²(x + 1) = x³ + x²; Subtract: 5x² + 11x + 6 Step 4: 5x² ÷ x = 5x Step 5: 5x(x + 1) = 5x² + 5x; Subtract: 6x + 6 Step 6: 6x ÷ x = 6 Step 7: 6(x + 1) = 6x + 6; Subtract: 0 Step 8: Quotient = x² + 5x + 6, Remainder = 0 Final Answer: x³ + 6x² + 11x + 6 = (x + 1)(x² + 5x + 6) = (x + 1)(x + 2)(x + 3)

Step 1: By remainder theorem, remainder = p(2) where p(x) = 2x³ - 3x² + x - 5 Step 2: p(2) = 2(2)³ - 3(2)² + 2 - 5 Step 3: = 2(8) - 3(4) + 2 - 5 Step 4: = 16 - 12 + 2 - 5 Step 5: = 1 Final Answer: Remainder = 1

Step 1: Compare with ax² + bx + c: a = 4, b = 8, c = 3 Step 2: Find two numbers that multiply to ac = 12 and add to b = 8 Step 3: Numbers are 6 and 2 (6 × 2 = 12, 6 + 2 = 8) Step 4: 4x² + 8x + 3 = 4x² + 6x + 2x + 3 Step 5: = 2x(2x + 3) + 1(2x + 3) Step 6: = (2x + 1)(2x + 3) Final Answer: (2x + 1)(2x + 3)

Step 1: Since x - 2 is a factor, p(2) = 0 Step 2: Divide x³ - 2x² - 5x + 6 by x - 2 Step 3: Using synthetic division or long division: Quotient = x² - 5 - 6/(x-2)... let me recalculate Step 4: x³ - 2x² - 5x + 6 = (x - 2)(x² + 0x - 5) + remainder Actually: divide properly Step 5: (x - 2)(x² - 3) = x³ - 3x - 2x² + 6 = x³ - 2x² - 3x + 6 (not matching) Step 6: Let me factor directly: (x - 2)(x² + ax + b) Expanding: x³ + ax² + bx - 2x² - 2ax - 2b = x³ + (a-2)x² + (b-2a)x - 2b Step 7: Comparing: a - 2…

Step 1: (3x + 4)² = (3x)² + 2(3x)(4) + 4² Step 2: = 9x² + 24x + 16 Final Answer: 9x² + 24x + 16