Polynomials — Telangana (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Polynomials" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Polynomials" — important questions with detailed answers, download PDF for board…
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Q1: State the relationship between zeroes and coefficients of a 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: 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 ✓
Q2: Define a polynomial. What do you mean by the degree of a polynomial?
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) …
Q3: Divide p(x) = x³ + 3x² + 3x + 1 by g(x) = x + 1 using long division.
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…
Q4: Find the zeroes of the polynomial p(x) = 2x² - 7x + 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 a…
Q5: If α and β are zeroes of x² - 4x + 3, find the polynomial whose zeroes are 2α and 2β.
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 …
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