Polynomials — Previous Year Questions (Class 10 Mathematics)
Polynomials explores zeros of polynomials, relationships between zeros and coefficients, and factorization techniques. This chapter is essential for solvin
TL;DR: Polynomials explores zeros of polynomials, relationships between zeros and coefficients, and factorization techniques. This chapter is essential for s…
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Polynomials explores zeros of polynomials, relationships between zeros and coefficients, and factorization techniques. This chapter is essential for solvin
Polynomials — Previous Year Questions with Solutions
Q (2023, 3 marks): If alpha and beta are zeros of polynomial x^2 - 5x + k, and alpha - beta = 1, find k.
Answer: Sum of zeros: alpha + beta = 5
Product of zeros: alpha × beta = k
Given: alpha - beta = 1
From alpha + beta = 5 and alpha - beta = 1:
Adding: 2*alpha = 6, so alpha = 3
Therefore beta = 2
k = alpha × beta = 3 × 2 = 6
Q (2022, 2 marks): Find a quadratic polynomial whose zeros are 2 and -3.
Answer: Sum of zeros = 2 + (-3) = -1
Product of zeros = 2 × (-3) = -6
Quadratic polynomial: x^2 - (sum)x + (product)
= x^2 - (-1)x + (-6)
= x^2 + x - 6
Verification: (x - 2)(x + 3) = x^2 + 3x - 2x - 6 = x^2 + x - 6 ✓
Q (2023, 3 marks): Divide 3x^3 - x^2 - 3x + 1 by x^2 - 1 and find quotient and remainder.
Answer: Using polynomial long division:
3x^3 - x^2 - 3x + 1 divided by x^2 - 1
First term of quotient: 3x^3 / x^2 = 3x
3x(x^2 - 1) = 3x^3 - 3x
Subtract: (3x^3 - x^2 - 3x + 1) - (3x^3 - 3x) = -x^2 + 1
Next term of quotient: -x^2 / x^2 = -1
-1(x^2 - 1) = -x^2 + 1
Subtract: (-x^2 + 1) - (-x^2 + 1) = 0
Quotient = 3x - 1, Remainder = 0
Q (2021, 3 marks): If one zero of the polynomial x^2 + (k+1)x + (k^2 - 5k + 6) is 2, find the other zero.
Answer: If 2 is a zero, then: (2)^2 + (k+1)(2) + (k^2 - 5k + 6) = 0
4 + 2k + 2 + k^2 - 5k + 6 = 0
k^2 - 3k + 12 = 0
Using discriminant: 9 - 48 = -39 < 0, no real solution
Let me verify problem: Assuming it's solvable, k^2 - 3k + 12 should give real k.
Actually, if 2 is a zero: 4 + 2(k+1) + (k^2 - 5k + 6) = 0
4 + 2k + 2 + k^2 - 5k + 6 = 0
k^2 - 3k + 12 = 0... This has no real solution.
Assuming corrected polynomial or value, use Vieta's: if alpha = 2, then
beta = (k^2 - 5k + 6)/2 once k is found.
Q (2022, 2 marks): Verify that 1, -1, and 2 are zeros of p(x) = x^3 - 2x^2 - x + 2.
Answer: p(1) = (1)^3 - 2(1)^2 - 1 + 2 = 1 - 2 - 1 + 2 = 0 ✓
p(-1) = (-1)^3 - 2(-1)^2 - (-1) + 2 = -1 - 2 + 1 + 2 = 0 ✓
p(2) = (2)^3 - 2(2)^2 - 2 + 2 = 8 - 8 - 2 + 2 = 0 ✓
All three values are zeros of p(x).
Q (2023, 3 marks): Find all zeros of p(x) = 2x^3 - 5x^2 - x + 6.
Answer: Test rational roots using factors of 6/2: ±1, ±2, ±3, ±6, ±1/2, ±3/2
p(1) = 2 - 5 - 1 + 6 = 2 ≠ 0
p(-1) = -2 - 5 + 1 + 6 = 0, so (x + 1) is a factor
Divide: 2x^3 - 5x^2 - x + 6 = (x + 1)(2x^2 - 7x + 6)
Factor 2x^2 - 7x + 6 = (2x - 3)(x - 2)
Zeros: x = -1, x = 3/2, x = 2
Frequently Asked Questions
What is the relationship between zeros and coefficients of a quadratic?
For quadratic ax^2 + bx + c, if alpha and beta are zeros, then alpha + beta = -b/a and alpha × beta = c/a.
How do you find zeros of a cubic polynomial?
Use the rational root theorem to test possible rational roots, then factor out linear terms to reduce to a quadratic, which can be solved using the quadratic formula.
More Class 10 Mathematics PYQs
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