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

Free step-by-step Telangana (SSC) Class 9 Mathematics solutions for "Polynomials and Factorisation" — important questions with detailed answers, download PDF for board exam preparation.

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

Free step-by-step Telangana (SSC) Class 9 Mathematics solutions for "Polynomials and Factorisation" — important questions with detailed answers, download PDF for board exam preparation.

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Polynomials and Factorisation — Textbook Questions Answered (8)

Q1: Divide x³ + 6x² + 11x + 6 by x + 1 using long division.

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)

Q2: Find remainder when 2x³ - 3x² + x - 5 is divided by x - 2 using remainder theorem.

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

Q3: Factorize 4x² + 8x + 3.

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)

Q4: If x - 2 is a factor of x³ - 2x² - 5x + 6, find the other factors.

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 = -2 → a = 0; -2b = 6 → b = -3 Step 8: (x - 2)(x² - 3) = (x - 2)(x - √3)(x + √3) Final Answer: Other factors are (x² - 3) or (x - √3)(x + √3)

Q5: Using identity (a + b)² = a² + 2ab + b², expand (3x + 4)².

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

Q6: Using identity a² - b² = (a + b)(a - b), factorize 25x² - 49.

Step 1: 25x² - 49 = (5x)² - 7² Step 2: = (5x + 7)(5x - 7) Final Answer: (5x + 7)(5x - 7)

Q7: Factorize x³ - 8.

Step 1: x³ - 8 = x³ - 2³ Step 2: Use identity a³ - b³ = (a - b)(a² + ab + b²) Step 3: = (x - 2)(x² + 2x + 4) Step 4: Check if x² + 2x + 4 factors further: discriminant = 4 - 16 = -12 < 0 Step 5: No real factors Final Answer: (x - 2)(x² + 2x + 4)

Q8: Find all zeros of p(x) = x³ - 6x² + 11x - 6.

Step 1: Test rational roots ±1, ±2, ±3, ±6 Step 2: p(1) = 1 - 6 + 11 - 6 = 0 ✓ So x = 1 is a zero Step 3: Divide by (x - 1): x³ - 6x² + 11x - 6 = (x - 1)(x² - 5x + 6) Step 4: Factorize x² - 5x + 6 = (x - 2)(x - 3) Step 5: p(x) = (x - 1)(x - 2)(x - 3) Final Answer: Zeros are x = 1, 2, 3

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