Polynomials — Andhra Pradesh (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: αβ = c/a These relationships work in both directions: given the coefficients, you can find the sum and product of the zeroes without solving the equation at all; conversely, given a desired sum and product, you can construct the corresponding quadratic directly as x² − (sum)x + (product) = 0.

A polynomial is an algebraic expression built from variables and constants using only addition, subtraction, and non-negative whole-number exponents of the variable — for example, 4x³ − x² + 2x − 9. The degree of a polynomial is the highest power of the variable that appears in it. In 4x³ − x² + 2x − 9, the highest power present is 3, so this is a degree-3 (cubic) polynomial. A polynomial's degree also tells you the MAXIMUM possible number of zeroes it can have — a degree-n polynomial has at most n zeroes.

Divide x³ + 5x² + 8x + 4 by x + 2. Step 1: x³ ÷ x = x². Multiply: x²(x + 2) = x³ + 2x². Subtract: (x³ + 5x²) − (x³ + 2x²) = 3x². Bring down 8x → 3x² + 8x. Step 2: 3x² ÷ x = 3x. Multiply: 3x(x + 2) = 3x² + 6x. Subtract: (3x² + 8x) − (3x² + 6x) = 2x. Bring down 4 → 2x + 4. Step 3: 2x ÷ x = 2. Multiply: 2(x + 2) = 2x + 4. Subtract: (2x + 4) − (2x + 4) = 0. Quotient: x² + 3x + 2, Remainder: 0 Since the remainder is 0, (x + 2) divides evenly into p(x), meaning it is an exact factor: p(x) = (x + 2)(x² + 3x + 2).

Using the quadratic formula: x = [−b ± √(b² − 4ac)] / 2a, with a = 2, b = −5, c = −3. Discriminant: b² − 4ac = (−5)² − 4(2)(−3) = 25 + 24 = 49 = 7² x = [5 ± 7] / 4 x = 12/4 = 3, or x = −2/4 = −1/2 Verification: p(3) = 2(9) − 5(3) − 3 = 18 − 15 − 3 = 0 ✓ p(−1/2) = 2(1/4) − 5(−1/2) − 3 = 1/2 + 5/2 − 3 = 3 − 3 = 0 ✓ The zeroes are 3 and −1/2.

For x² − 6x + 8: sum of zeroes α + β = 6, product αβ = 8. (In fact α = 2, β = 4, since 2 + 4 = 6 and 2 × 4 = 8.) For the new zeroes 2α and 2β: New sum = 2α + 2β = 2(α + β) = 2(6) = 12 New product = (2α)(2β) = 4αβ = 4(8) = 32 The required polynomial: x² − (sum)x + (product) = x² − 12x + 32 Verification: the new zeroes are 2(2) = 4 and 2(4) = 8. Sum = 12 ✓, product = 32 ✓.

By the Remainder Theorem, dividing p(x) by (x − a) leaves a remainder equal to p(a). Here a = 3, so the remainder = p(3). p(3) = (3)³ − 2(3)² + 4(3) − 3 = 27 − 18 + 12 − 3 = 18 The remainder is 18.

By the Factor Theorem, (x − a) is a factor of p(x) if and only if p(a) = 0. Here a = 4: p(4) = (4)³ − 9(4)² + 26(4) − 24 = 64 − 144 + 104 − 24 = 0 Since p(4) = 0, (x − 4) IS a factor of p(x). (In fact p(x) = (x − 2)(x − 3)(x − 4), so the other two factors are (x − 2) and (x − 3).)

If the zeroes are r₁, r₂, r₃, the cubic is (x − r₁)(x − r₂)(x − r₃). Here r₁ = 1, r₂ = −2, r₃ = 5: p(x) = (x − 1)(x + 2)(x − 5) Expand (x − 1)(x + 2) first: x² + x − 2 Now multiply by (x − 5): (x² + x − 2)(x − 5) = x³ − 5x² + x² − 5x − 2x + 10 = x³ − 4x² − 7x + 10 The required cubic polynomial is p(x) = x³ − 4x² − 7x + 10.

Factorise the numerator x² − 7x + 12. We need two numbers that multiply to 12 and add to −7: these are −3 and −4. x² − 7x + 12 = (x − 3)(x − 4) So (x² − 7x + 12)/(x − 4) = (x − 3)(x − 4)/(x − 4) = x − 3, for x ≠ 4. The simplified expression is x − 3.