Polynomials — Telangana (SSC) Class 10 Mathematics Solutions (Free)

For a quadratic polynomial ax² + bx + c with zeroes α and β: Sum of zeroes: α + β = -b/a Product of zeroes: α × β = c/a Example: For the polynomial 2x² - 5x + 3, with a = 2, b = -5, c = 3: If the zeroes are α and β, then: α + β = -(-5)/2 = 5/2 α × β = 3/2 We can verify: zeroes are 1 and 3/2 1 + 3/2 = 5/2 ✓ 1 × 3/2 = 3/2 ✓

Polynomial: A polynomial is an algebraic expression consisting of terms with variables and coefficients, where variables have non-negative integer powers. General form: p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ where aₙ, aₙ₋₁, ..., a₁, a₀ are constants and n is a non-negative integer. Degree of a polynomial: The degree of a polynomial is the highest power of the variable in the polynomial. Examples: - p(x) = 3x² + 2x + 5 has degree 2 (quadratic) - p(x) = x³ - 4x + 1 has degree 3 (cubic) - p(x) = 7x + 2 has degree 1 (linear) - p(x) = 5 has degree 0 (constant)

Polynomial long division of (x³ + 3x² + 3x + 1) ÷ (x + 1): Step 1: Divide first term of dividend by first term of divisor x³ ÷ x = x² Multiply: x²(x + 1) = x³ + x² Subtract: (x³ + 3x² + 3x + 1) - (x³ + x²) = 2x² + 3x + 1 Step 2: Repeat with new dividend 2x² + 3x + 1 2x² ÷ x = 2x Multiply: 2x(x + 1) = 2x² + 2x Subtract: (2x² + 3x + 1) - (2x² + 2x) = x + 1 Step 3: Repeat with new dividend x + 1 x ÷ x = 1 Multiply: 1(x + 1) = x + 1 Subtract: (x + 1) - (x + 1) = 0 Quotient: x² + 2x + 1 Remainder: 0 Therefore: (x³ + 3x² + 3x + 1) = (x + 1)(x² + 2x + 1) Note: x² + 2x + 1 = (x + 1)², so the answer is (x + 1)³

Using the factorisation method: We need to find factors of 2 × 3 = 6 that add up to -7. The factors are -6 and -1 (since -6 + (-1) = -7 and -6 × (-1) = 6) 2x² - 7x + 3 = 2x² - 6x - x + 3 = 2x(x - 3) - 1(x - 3) = (2x - 1)(x - 3) Setting each factor to zero: 2x - 1 = 0 → x = 1/2 x - 3 = 0 → x = 3 Verification: For x = 1/2: 2(1/2)² - 7(1/2) + 3 = 2(1/4) - 7/2 + 3 = 1/2 - 7/2 + 3 = 0 ✓ For x = 3: 2(3)² - 7(3) + 3 = 18 - 21 + 3 = 0 ✓ Therefore, the zeroes are 1/2 and 3

Step 1: Find the zeroes of x² - 4x + 3 Factorising: x² - 4x + 3 = (x - 1)(x - 3) Zeroes: α = 1, β = 3 Step 2: Find 2α and 2β 2α = 2(1) = 2 2β = 2(3) = 6 Step 3: Method 1 - Form polynomial with new zeroes Sum of new zeroes: 2α + 2β = 2 + 6 = 8 Product of new zeroes: (2α)(2β) = 2 × 6 = 12 Polynomial: x² - (sum)x + (product) = x² - 8x + 12 Step 4: Verification For x = 2: (2)² - 8(2) + 12 = 4 - 16 + 12 = 0 ✓ For x = 6: (6)² - 8(6) + 12 = 36 - 48 + 12 = 0 ✓ Therefore, the required polynomial is x² - 8x + 12

Remainder Theorem: When a polynomial p(x) is divided by (x - a), the remainder is p(a). Here, p(x) = 2x³ - 3x² + 7x - 5 and we're dividing by (x - 2), so a = 2. Remainder = p(2) = 2(2)³ - 3(2)² + 7(2) - 5 = 2(8) - 3(4) + 14 - 5 = 16 - 12 + 14 - 5 = 13 Therefore, the remainder is 13

Using the Factor Theorem: (x - a) is a factor of p(x) if and only if p(a) = 0. To check if (x - 1) is a factor, calculate p(1): p(1) = (1)³ - 6(1)² + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 Since p(1) = 0, (x - 1) is a factor of p(x). Additionally, we can factor completely: p(x) = x³ - 6x² + 11x - 6 We know (x - 1) is a factor, so we divide: p(x) = (x - 1)(x² - 5x + 6) = (x - 1)(x - 2)(x - 3) Verification: The zeroes are 1, 2, and 3, and p(1) = 0 ✓

If α = 2, β = -1, and γ = 3 are the zeroes of a cubic polynomial, then: The polynomial can be written as: p(x) = k(x - α)(x - β)(x - γ) p(x) = k(x - 2)(x + 1)(x - 3) where k is any non-zero constant. Taking k = 1: p(x) = (x - 2)(x + 1)(x - 3) Expanding step by step: First, multiply (x - 2)(x + 1): (x - 2)(x + 1) = x² + x - 2x - 2 = x² - x - 2 Now multiply (x² - x - 2)(x - 3): = x³ - 3x² - x² + 3x - 2x + 6 = x³ - 4x² + x + 6 Verification: p(2) = 8 - 16 + 2 + 6 = 0 ✓ p(-1) = -1 - 4 - 1 + 6 = 0 ✓ p(3) = 27 - 36 + 3 + 6 = 0 ✓ Therefore, the polynomial is x³ - 4x² + x + 6

Simplify (x² - 5x + 6)/(x - 2) Step 1: Factorise the numerator x² - 5x + 6 We need two numbers that multiply to 6 and add to -5. The factors are -2 and -3 (since (-2) × (-3) = 6 and (-2) + (-3) = -5) x² - 5x + 6 = (x - 2)(x - 3) Step 2: Substitute the factorisation into the expression (x² - 5x + 6)/(x - 2) = ((x - 2)(x - 3))/(x - 2) Step 3: Cancel the common factor (x - 2) Note: This is valid for x ≠ 2 = (x - 3) Therefore, (x² - 5x + 6)/(x - 2) = x - 3 (for x ≠ 2)