Polynomials — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)

Free step-by-step Karnataka (SSLC) Class 10 Mathematics solutions for "Polynomials" — important questions with detailed answers, download PDF for board exam preparation.

Polynomials — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)

Free step-by-step Karnataka (SSLC) Class 10 Mathematics solutions for "Polynomials" — important questions with detailed answers, download PDF for board exam preparation.

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TL;DR: Free step-by-step Karnataka (SSLC) Class 10 Mathematics solutions for "Polynomials" — important questions with detailed answers, download PDF for boar…

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Polynomials — Textbook Questions Answered (9)

Q1: State the relationship between the zeroes and coefficients of the quadratic polynomial ax² + bx + c.

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 let you find the sum and product of the roots directly from the coefficients without solving the equation, and conversely let you build a quadratic once you know its sum and product of zeroes: x² − (sum)x + (product) = 0.

Q2: What is a polynomial, and what is meant by its degree? Give an example of a degree-3 polynomial.

A polynomial is an expression made up of variables and constants, combined using only addition, subtraction, and non-negative whole-number powers of the variable — for example 5x³ − 2x² + x − 7. The degree of a polynomial is the highest power of the variable appearing in it. In 5x³ − 2x² + x − 7, the highest power is 3, so this is a degree-3 (cubic) polynomial. A polynomial of degree n has at most n zeroes.

Q3: Divide p(x) = x³ + 4x² + 5x + 2 by g(x) = x + 2 using long division.

Divide x³ + 4x² + 5x + 2 by x + 2. Step 1: x³ ÷ x = x². Multiply: x²(x + 2) = x³ + 2x². Subtract: (x³ + 4x²) − (x³ + 2x²) = 2x². Bring down 5x → 2x² + 5x. Step 2: 2x² ÷ x = 2x. Multiply: 2x(x + 2) = 2x² + 4x. Subtract: (2x² + 5x) − (2x² + 4x) = x. Bring down 2 → x + 2. Step 3: x ÷ x = 1. Multiply: 1(x + 2) = x + 2. Subtract: (x + 2) − (x + 2) = 0. Quotient: x² + 2x + 1, Remainder: 0 So p(x) = (x + 2)(x² + 2x + 1), and since the remainder is 0, (x + 2) is an exact factor.

Q4: Find the zeroes of the polynomial p(x) = 3x² − 11x + 6.

Using the quadratic formula: x = [−b ± √(b² − 4ac)] / 2a, with a = 3, b = −11, c = 6. Discriminant: b² − 4ac = (−11)² − 4(3)(6) = 121 − 72 = 49 = 7² x = [11 ± 7] / 6 x = 18/6 = 3, or x = 4/6 = 2/3 Verification: p(3) = 3(9) − 11(3) + 6 = 27 − 33 + 6 = 0 ✓ p(2/3) = 3(4/9) − 11(2/3) + 6 = 4/3 − 22/3 + 18/3 = 0 ✓ The zeroes are 3 and 2/3.

Q5: If α and β are the zeroes of x² − 5x + 6, find the polynomial whose zeroes are 3α and 3β.

For x² − 5x + 6: sum of zeroes α + β = 5, product αβ = 6. (In fact α = 2, β = 3, since 2 + 3 = 5 and 2 × 3 = 6.) For the new zeroes 3α and 3β: New sum = 3α + 3β = 3(α + β) = 3(5) = 15 New product = (3α)(3β) = 9αβ = 9(6) = 54 The required polynomial: x² − (sum)x + (product) = x² − 15x + 54 Verification: the new zeroes are 3(2) = 6 and 3(3) = 9. Sum = 15 ✓, product = 54 ✓.

Q6: Use the Remainder Theorem to find the remainder when p(x) = 3x³ − 4x² + 5x − 8 is divided by (x − 2).

By the Remainder Theorem, dividing p(x) by (x − a) leaves a remainder equal to p(a). Here a = 2, so the remainder = p(2). p(2) = 3(2)³ − 4(2)² + 5(2) − 8 = 3(8) − 4(4) + 10 − 8 = 24 − 16 + 10 − 8 = 10 The remainder is 10.

Q7: Determine if (x − 2) is a factor of p(x) = x³ − 7x² + 14x − 8.

By the Factor Theorem, (x − a) is a factor of p(x) if and only if p(a) = 0. Here a = 2: p(2) = (2)³ − 7(2)² + 14(2) − 8 = 8 − 28 + 28 − 8 = 0 Since p(2) = 0, (x − 2) IS a factor of p(x). (In fact p(x) = (x − 1)(x − 2)(x − 4), so the other two factors are (x − 1) and (x − 4).)

Q8: Find a cubic polynomial whose zeroes are 2, −1, and 3.

If the zeroes are r₁, r₂, r₃, the cubic is (x − r₁)(x − r₂)(x − r₃). Here r₁ = 2, r₂ = −1, r₃ = 3: p(x) = (x − 2)(x + 1)(x − 3) Expand (x − 2)(x + 1) first: x² − x − 2 Now multiply by (x − 3): (x² − x − 2)(x − 3) = x³ − 3x² − x² + 3x − 2x + 6 = x³ − 4x² + x + 6 The required cubic polynomial is p(x) = x³ − 4x² + x + 6.

Q9: Simplify (x² − 5x + 6)/(x − 3) by factorisation.

Factorise the numerator x² − 5x + 6. We need two numbers that multiply to 6 and add to −5: these are −2 and −3. x² − 5x + 6 = (x − 2)(x − 3) So (x² − 5x + 6)/(x − 3) = (x − 2)(x − 3)/(x − 3) = x − 2, for x ≠ 3. The simplified expression is x − 2.

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