📊 Polynomials Question Paper
80 Marks | 2 Hours | Class 9
Section A: 10 × 1 Mark = 10 Marks
Section B: 10 × 2 Marks = 20 Marks
Section C: 10 × 5 Marks = 50 Marks
Total = 80 Marks
Section B: 10 × 2 Marks = 20 Marks
Section C: 10 × 5 Marks = 50 Marks
Total = 80 Marks
SECTION A: Multiple Choice (1 Mark Each)
Q1. Which of the following is a polynomial? 1M
a) x + 1/x b) √x + 2 c) 2x² + 5x - 3 d) x⁻²
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✓ Answer: c) 2x² + 5x - 3
Solution: A polynomial must have whole number exponents. Only option c has exponents 2, 1, 0 (all whole numbers).
Solution: A polynomial must have whole number exponents. Only option c has exponents 2, 1, 0 (all whole numbers).
Q2. What is the degree of polynomial 5x⁶ - 4x² - 6? 1M
a) 2 b) 6 c) 4 d) 5
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✓ Answer: b) 6
Solution: Degree = Highest power of variable = 6
Solution: Degree = Highest power of variable = 6
Q3. Polynomial with two terms is called? 1M
a) Monomial b) Binomial c) Trinomial d) None
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✓ Answer: b) Binomial
Solution: Binomial = 2 terms (bi = two)
Solution: Binomial = 2 terms (bi = two)
Q4. If p(x) = 2x + 1, find p(-1/2)? 1M
a) 1 b) 0 c) -1 d) 2
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✓ Answer: b) 0
Solution: p(-1/2) = 2(-1/2) + 1 = -1 + 1 = 0
Solution: p(-1/2) = 2(-1/2) + 1 = -1 + 1 = 0
Q5. Zero of p(x) = 3x - 6? 1M
a) 3 b) 6 c) 2 d) -2
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✓ Answer: c) 2
Solution: Set 3x - 6 = 0, so 3x = 6, therefore x = 2
Solution: Set 3x - 6 = 0, so 3x = 6, therefore x = 2
Q6. Which is a linear polynomial? 1M
a) x² + 5 b) 4x + 7 c) x³ - 2 d) 8
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✓ Answer: b) 4x + 7
Solution: Linear polynomial has degree 1. Only 4x + 7 has degree 1.
Solution: Linear polynomial has degree 1. Only 4x + 7 has degree 1.
Q7. How many zeroes does a linear polynomial have? 1M
a) 0 b) 1 c) 2 d) Infinite
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✓ Answer: b) 1
Solution: Every linear polynomial has exactly ONE zero.
Solution: Every linear polynomial has exactly ONE zero.
Q8. (x + y)² = ? 1M
a) x² + y² b) x² + 2xy + y² c) x² - 2xy + y² d) 2x + 2y
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✓ Answer: b) x² + 2xy + y²
Solution: This is Identity I: (x + y)² = x² + 2xy + y²
Solution: This is Identity I: (x + y)² = x² + 2xy + y²
Q9. x² - y² = ? 1M
a) (x+y)² b) (x-y)² c) (x+y)(x-y) d) x²+2xy+y²
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✓ Answer: c) (x+y)(x-y)
Solution: This is Difference of Squares: x² - y² = (x + y)(x - y)
Solution: This is Difference of Squares: x² - y² = (x + y)(x - y)
Q10. If (x - 2) is a factor of p(x), then p(2) = ? 1M
a) 1 b) -2 c) 0 d) 2
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✓ Answer: c) 0
Solution: By Factor Theorem: If (x - a) is a factor, then p(a) = 0. So p(2) = 0.
Solution: By Factor Theorem: If (x - a) is a factor, then p(a) = 0. So p(2) = 0.
SECTION B: Short Answer (2 Marks Each)
Q11. Find value of p(x) = x² - 3x + 2 at x = 2 2M
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Solution: p(2) = (2)² - 3(2) + 2 = 4 - 6 + 2 = 0
Q12. Is -1 a zero of p(x) = x³ + x² + x + 1? 2M
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Solution: p(-1) = (-1)³ + (-1)² + (-1) + 1 = -1 + 1 - 1 + 1 = 0. Yes, -1 is a zero.
Q13. Find zero of p(x) = 2x - 5 2M
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Solution: 2x - 5 = 0 → 2x = 5 → x = 5/2
Q14. Classify 3x² + 2x + 1 2M
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Solution: By degree: Quadratic (degree 2). By terms: Trinomial (3 terms).
Q15. Coefficient of x in 5x³ + 4x² - 3x + 7 2M
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Solution: Coefficient of x = -3
Q16. Degree of y⁵ + y³ - 5y + 2 2M
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Solution: Highest power = 5. Degree = 5
Q17. Expand (x + 3)² 2M
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Solution: (x + 3)² = x² + 6x + 9
Q18. Factor x² - 9 2M
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Solution: x² - 9 = (x + 3)(x - 3)
Q19. Is (x+2) factor of x² + 5x + 6? 2M
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Solution: p(-2) = 4 - 10 + 6 = 0. Yes, (x+2) is a factor.
Q20. Find p(0) + p(1) if p(x) = x² - 2x + 3 2M
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Solution: p(0) = 3, p(1) = 1 - 2 + 3 = 2. So p(0) + p(1) = 5
SECTION C: Long Answer (5 Marks Each)
Q21. Using Factor Theorem, check whether (x + 2) is a factor of x³ + 3x² + 5x + 6 5M
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Solution:
Step 1: Zero of (x + 2) is -2
Step 2: Calculate p(-2) = (-2)³ + 3(-2)² + 5(-2) + 6
Step 3: = -8 + 12 - 10 + 6 = 0
Step 4: Since p(-2) = 0, (x + 2) IS a factor
Step 5: By division: x³ + 3x² + 5x + 6 = (x + 2)(x² + x + 3)
Step 1: Zero of (x + 2) is -2
Step 2: Calculate p(-2) = (-2)³ + 3(-2)² + 5(-2) + 6
Step 3: = -8 + 12 - 10 + 6 = 0
Step 4: Since p(-2) = 0, (x + 2) IS a factor
Step 5: By division: x³ + 3x² + 5x + 6 = (x + 2)(x² + x + 3)
Q22. Factorize 6x² + 17x + 5 by splitting the middle term 5M
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Solution:
Step 1: Find p and q such that p + q = 17 and pq = 6 × 5 = 30
Step 2: Pairs of 30: 1×30, 2×15, 3×10, 5×6
Step 3: We need 2 + 15 = 17 ✓
Step 4: 6x² + 17x + 5 = 6x² + 2x + 15x + 5
Step 5: = 2x(3x + 1) + 5(3x + 1)
Step 6: = (3x + 1)(2x + 5)
Answer: (3x + 1)(2x + 5)
Step 1: Find p and q such that p + q = 17 and pq = 6 × 5 = 30
Step 2: Pairs of 30: 1×30, 2×15, 3×10, 5×6
Step 3: We need 2 + 15 = 17 ✓
Step 4: 6x² + 17x + 5 = 6x² + 2x + 15x + 5
Step 5: = 2x(3x + 1) + 5(3x + 1)
Step 6: = (3x + 1)(2x + 5)
Answer: (3x + 1)(2x + 5)
Q23. Expand (2x - 3y + 4z)² using algebraic identity 5M
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Solution:
Use Identity V: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Where a = 2x, b = -3y, c = 4z
Step 1: (2x)² = 4x²
Step 2: (-3y)² = 9y²
Step 3: (4z)² = 16z²
Step 4: 2(2x)(-3y) = -12xy
Step 5: 2(-3y)(4z) = -24yz
Step 6: 2(2x)(4z) = 16xz
Answer: 4x² + 9y² + 16z² - 12xy - 24yz + 16xz
Use Identity V: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Where a = 2x, b = -3y, c = 4z
Step 1: (2x)² = 4x²
Step 2: (-3y)² = 9y²
Step 3: (4z)² = 16z²
Step 4: 2(2x)(-3y) = -12xy
Step 5: 2(-3y)(4z) = -24yz
Step 6: 2(2x)(4z) = 16xz
Answer: 4x² + 9y² + 16z² - 12xy - 24yz + 16xz
Q24. Find all zeroes of polynomial p(x) = x² - 5x + 6 5M
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Solution:
Step 1: Factorize x² - 5x + 6
Step 2: Find p + q = -5, pq = 6
Step 3: Numbers are -2 and -3
Step 4: x² - 5x + 6 = (x - 2)(x - 3)
Step 5: Set each factor to 0:
x - 2 = 0 → x = 2
x - 3 = 0 → x = 3
Answer: Zeroes are 2 and 3
Step 1: Factorize x² - 5x + 6
Step 2: Find p + q = -5, pq = 6
Step 3: Numbers are -2 and -3
Step 4: x² - 5x + 6 = (x - 2)(x - 3)
Step 5: Set each factor to 0:
x - 2 = 0 → x = 2
x - 3 = 0 → x = 3
Answer: Zeroes are 2 and 3
Q25. Factorize 8x³ + 27y³ + 36x²y + 54xy² 5M
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Solution:
Step 1: Rewrite as (2x)³ + (3y)³ + 3(4x²)(3y) + 3(2x)(9y²)
Step 2: = (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)²
Step 3: Recognize pattern from Identity VI: (a + b)³ = a³ + b³ + 3a²b + 3ab²
Step 4: Where a = 2x, b = 3y
Step 5: = (2x + 3y)³
Answer: (2x + 3y)³ or (2x + 3y)(2x + 3y)(2x + 3y)
Step 1: Rewrite as (2x)³ + (3y)³ + 3(4x²)(3y) + 3(2x)(9y²)
Step 2: = (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)²
Step 3: Recognize pattern from Identity VI: (a + b)³ = a³ + b³ + 3a²b + 3ab²
Step 4: Where a = 2x, b = 3y
Step 5: = (2x + 3y)³
Answer: (2x + 3y)³ or (2x + 3y)(2x + 3y)(2x + 3y)
Q26. If (x - 1) is a factor of kx² - 3x + k, find the value of k 5M
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Solution:
Step 1: By Factor Theorem, if (x - 1) is a factor, then p(1) = 0
Step 2: Substitute x = 1 in p(x) = kx² - 3x + k
Step 3: p(1) = k(1)² - 3(1) + k = 0
Step 4: k - 3 + k = 0
Step 5: 2k - 3 = 0
Step 6: 2k = 3
Answer: k = 3/2 or 1.5
Step 1: By Factor Theorem, if (x - 1) is a factor, then p(1) = 0
Step 2: Substitute x = 1 in p(x) = kx² - 3x + k
Step 3: p(1) = k(1)² - 3(1) + k = 0
Step 4: k - 3 + k = 0
Step 5: 2k - 3 = 0
Step 6: 2k = 3
Answer: k = 3/2 or 1.5
Q27. Factorize x³ - 23x² + 142x - 120 completely 5M
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Solution:
Step 1: Test divisors of 120: ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±24, ±30, ±60, ±120
Step 2: p(1) = 1 - 23 + 142 - 120 = 0, so (x - 1) is a factor
Step 3: Divide: x³ - 23x² + 142x - 120 = (x - 1)(x² - 22x + 120)
Step 4: Factorize x² - 22x + 120
Step 5: Need p + q = -22, pq = 120
Step 6: Numbers are -10 and -12
Step 7: x² - 22x + 120 = (x - 10)(x - 12)
Answer: (x - 1)(x - 10)(x - 12)
Step 1: Test divisors of 120: ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±24, ±30, ±60, ±120
Step 2: p(1) = 1 - 23 + 142 - 120 = 0, so (x - 1) is a factor
Step 3: Divide: x³ - 23x² + 142x - 120 = (x - 1)(x² - 22x + 120)
Step 4: Factorize x² - 22x + 120
Step 5: Need p + q = -22, pq = 120
Step 6: Numbers are -10 and -12
Step 7: x² - 22x + 120 = (x - 10)(x - 12)
Answer: (x - 1)(x - 10)(x - 12)
Q28. Use algebraic identity to evaluate 105 × 106 without direct multiplication 5M
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Solution:
Step 1: Rewrite: 105 × 106 = (100 + 5)(100 + 6)
Step 2: Use Identity IV: (x + a)(x + b) = x² + (a + b)x + ab
Step 3: Where x = 100, a = 5, b = 6
Step 4: = (100)² + (5 + 6)(100) + (5)(6)
Step 5: = 10000 + 1100 + 30
Step 6: = 11130
Answer: 105 × 106 = 11,130
Step 1: Rewrite: 105 × 106 = (100 + 5)(100 + 6)
Step 2: Use Identity IV: (x + a)(x + b) = x² + (a + b)x + ab
Step 3: Where x = 100, a = 5, b = 6
Step 4: = (100)² + (5 + 6)(100) + (5)(6)
Step 5: = 10000 + 1100 + 30
Step 6: = 11130
Answer: 105 × 106 = 11,130
Q29. Factorize 4x² + 9y² + 16z² + 12xy - 24yz - 16xz 5M
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Solution:
Step 1: Rewrite: = (2x)² + (-3y)² + (4z)² + 2(2x)(-3y) + 2(-3y)(4z) + 2(2x)(4z)
Step 2: Recognize as (a + b + c)² form where a = 2x, b = -3y, c = 4z
Step 3: Use Identity V: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Step 4: = (2x - 3y + 4z)²
Answer: (2x - 3y + 4z)² or (2x - 3y + 4z)(2x - 3y + 4z)
Step 1: Rewrite: = (2x)² + (-3y)² + (4z)² + 2(2x)(-3y) + 2(-3y)(4z) + 2(2x)(4z)
Step 2: Recognize as (a + b + c)² form where a = 2x, b = -3y, c = 4z
Step 3: Use Identity V: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Step 4: = (2x - 3y + 4z)²
Answer: (2x - 3y + 4z)² or (2x - 3y + 4z)(2x - 3y + 4z)
Q30. Verify: x³ + y³ = (x + y)(x² - xy + y²) and factorize 8x³ + 125 5M
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Verification:
Step 1: Expand (x + y)(x² - xy + y²)
Step 2: = x(x² - xy + y²) + y(x² - xy + y²)
Step 3: = x³ - x²y + xy² + x²y - xy² + y³
Step 4: = x³ + y³ ✓ (Verified)
Factorization of 8x³ + 125:
Step 5: Rewrite as (2x)³ + (5)³
Step 6: Apply verified identity with a = 2x, b = 5
Step 7: = (2x + 5)[(2x)² - (2x)(5) + (5)²]
Step 8: = (2x + 5)(4x² - 10x + 25)
Answer: (2x + 5)(4x² - 10x + 25)
Step 1: Expand (x + y)(x² - xy + y²)
Step 2: = x(x² - xy + y²) + y(x² - xy + y²)
Step 3: = x³ - x²y + xy² + x²y - xy² + y³
Step 4: = x³ + y³ ✓ (Verified)
Factorization of 8x³ + 125:
Step 5: Rewrite as (2x)³ + (5)³
Step 6: Apply verified identity with a = 2x, b = 5
Step 7: = (2x + 5)[(2x)² - (2x)(5) + (5)²]
Step 8: = (2x + 5)(4x² - 10x + 25)
Answer: (2x + 5)(4x² - 10x + 25)
Total: 80 Marks | 30 Questions | Complete Solution Set
All questions with detailed solutions for Polynomials Chapter
All questions with detailed solutions for Polynomials Chapter
