CBSE Class 10 Mathematics Polynomials MCQs Set 26

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 02 Polynomials

Review structured MCQ sets for Class 10 Mathematics Chapter 02 Polynomials. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Chapter 02 Polynomials MCQs for Class 10 Mathematics

Navigate directly to the 50 objective questions for Chapter 02 Polynomials using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The product (x + a)(x + b) equals
(a) x² + (a + b)x + ab
(b) x² − (a − b)x + ab
(c) a² + (a − b)x + ab
(d) x² + (a − b)x − ab
Show Answer & Explanation

Answer: (a) x² + (a + b)x + ab

Explanation:
1. x(x + b) + a(x + b) = x² + bx + ax + ab = x² + (a + b)x + ab.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 2: The value of 150 × 98 is
(a) 10047
(b) 14700
(c) 14800
(d) 10470
Show Answer & Explanation

Answer: (b) 14700

Explanation:
1. 150 × 98 = 150(100 − 2) = 15000 − 300 = 14700.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 3: The value of (x + 2y + 2z)² + (x − 2y − 2z)² is
(a) 2x² + 8y² + 8z²
(b) 2x² + 8y² + 8z² + 8xyz
(c) 2x² + 8y² + 8z² + 16yz
(d) 2x² + 8y² + 8z² − 8yz
Show Answer & Explanation

Answer: (c) 2x² + 8y² + 8z² + 16yz

Explanation:
1. Put p = 2y + 2z: (x + p)² + (x − p)² = 2x² + 2p² = 2x² + 2(2y + 2z)² = 2x² + 8y² + 8z² + 16yz.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 4: The value of 25x² + 16y² + 40xy at x = 1 and y = −1 is
(a) 1
(b) 81
(c) −49
(d) 9
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. 25x² + 16y² + 40xy = (5x + 4y)². At x = 1, y = −1: (5 − 4)² = 1.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 5: The value of (a³ + b³ + c³ − 3abc)/(ab + bc + ca − a² − b² − c²) for a = −5, b = −6, c = 10 is
(a) −1
(b) 2
(c) 1
(d) −2
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. The numerator is (a + b + c)(a² + b² + c² − ab − bc − ca) and the denominator is −(a² + b² + c² − ab − bc − ca). So the ratio is −(a + b + c) = −(−5 − 6 + 10) = 1.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 6: If x + y + z = 1, xy + yz + zx = −1 and xyz = −1, then x³ + y³ + z³ equals
(a) 1
(b) −1
(c) 2
(d) −2
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. x² + y² + z² = 1 − 2(−1) = 3. x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx) = 1(3 + 1) = 4. So x³ + y³ + z³ = 4 + 3(−1) = 1.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 7: The degree of the zero polynomial is
(a) 0
(b) 1
(c) both 0 and 1
(d) not defined
Show Answer & Explanation

Answer: (d) not defined

Explanation:
1. The zero polynomial has all coefficients zero, so no highest power of x with non-zero coefficient exists; its degree is not defined.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 8: If x = 152 and y = −91, then 9x² + 30xy + 25y² equals
(a) 3600
(b) 5
(c) 0
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. 9x² + 30xy + 25y² = (3x + 5y)² = (456 − 455)² = 1.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 9: (2x − 1/x)² equals
(a) 4x² − 2 + 1/x²
(b) 4x² + 4 + 1/x²
(c) 2x² − 4 + 1/x²
(d) 4x² − 4 + 1/x²
Show Answer & Explanation

Answer: (d) 4x² − 4 + 1/x²

Explanation:
1. (2x − 1/x)² = 4x² − 2(2x)(1/x) + 1/x² = 4x² − 4 + 1/x².

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 10: If x + y = 3 and xy = −18, then x³ + y³ equals
(a) −135
(b) 27
(c) 189
(d) 81
Show Answer & Explanation

Answer: (c) 189

Explanation:
1. x³ + y³ = (x + y)³ − 3xy(x + y) = 27 − 3(−18)(3) = 27 + 162 = 189.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 11: The value of (1/2)³ + (1/3)³ − (5/6)³ is
(a) −5/12
(b) 5/12
(c) 0
(d) −5/6
Show Answer & Explanation

Answer: (a) −5/12

Explanation:
1. 1/2 + 1/3 − 5/6 = 0, so the sum of cubes = 3abc = 3(1/2)(1/3)(−5/6) = −5/12.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 12: The value of (0.2)³ − (0.3)³ + (0.1)³ is
(a) 0.018
(b) 0
(c) −0.018
(d) −0.18
Show Answer & Explanation

Answer: (c) −0.018

Explanation:
1. 0.2 − 0.3 + 0.1 = 0, so the value = 3(0.2)(−0.3)(0.1) = −0.018.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 13: The product (p² + 16)(p² − 1/4) equals
(a) p⁴ + (65/4)p² − 4
(b) p⁴ + (63/4)p² − 4
(c) p⁴ + (63/4)p² + 4
(d) p⁴ − (63/4)p² − 4
Show Answer & Explanation

Answer: (b) p⁴ + (63/4)p² − 4

Explanation:
1. With y = p²: (y + 16)(y − 1/4) = y² + (16 − 1/4)y − 4 = p⁴ + (63/4)p² − 4.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 14: The value of 102 × 106 is
(a) 10812
(b) 10712
(c) 10802
(d) 10612
Show Answer & Explanation

Answer: (a) 10812

Explanation:
1. 102 × 106 = (100 + 2)(100 + 6) = 10000 + 800 + 12 = 10812.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 15: The value of 34 × 36 is
(a) 1234
(b) 1224
(c) 1214
(d) 1296
Show Answer & Explanation

Answer: (b) 1224

Explanation:
1. 34 × 36 = (35 − 1)(35 + 1) = 1225 − 1 = 1224.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 16: Factorise 64a³ − 27b³ − 144a²b + 108ab².
(a) (4a − 3b)³
(b) (4a + 3b)³
(c) (4a − 3b)(16a² + 12ab + 9b²)
(d) (4a − 3b)²(4a + 3b)
Show Answer & Explanation

Answer: (a) (4a − 3b)³

Explanation:
1. This is (4a)³ − 3(4a)²(3b) + 3(4a)(3b)² − (3b)³ = (4a − 3b)³.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 17: Factorise x⁴ + 2x³y − 2xy³ − y⁴.
(a) (x − y)²(x + y)²
(b) (x − y)(x + y)³
(c) (x² + y²)(x + y)²
(d) (x − y)³(x + y)
Show Answer & Explanation

Answer: (b) (x − y)(x + y)³

Explanation:
1. x⁴ − y⁴ + 2xy(x² − y²) = (x² − y²)(x² + y² + 2xy) = (x − y)(x + y)(x + y)² = (x − y)(x + y)³.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 18: For distinct a, b, c the value of (a − b)²/((b − c)(c − a)) + (b − c)²/((a − b)(c − a)) + (c − a)²/((a − b)(b − c)) is
(a) 3
(b) 0
(c) 1
(d) a + b + c
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Put p = a − b, q = b − c, r = c − a with p + q + r = 0. The sum is (p³ + q³ + r³)/(pqr) = 3pqr/pqr = 3.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 19: If x² + 1/x² = 23, then x + 1/x equals (x > 0)
(a) √21
(b) 5
(c) −5
(d) 25
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. (x + 1/x)² = x² + 2 + 1/x² = 25, so x + 1/x = 5 for x > 0.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 20: The value of (107)² is
(a) 11409
(b) 11449
(c) 11349
(d) 11249
Show Answer & Explanation

Answer: (b) 11449

Explanation:
1. (100 + 7)² = 10000 + 1400 + 49 = 11449.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 21: The value of (1005)³ is
(a) 1015075000
(b) 1015050125
(c) 1015000125
(d) 1015075125
Show Answer & Explanation

Answer: (d) 1015075125

Explanation:
1. (1000 + 5)³ = 10⁹ + 3(10⁶)(5) + 3(1000)(25) + 125 = 1000000000 + 15000000 + 75000 + 125 = 1015075125.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 22: The value of (997)³ is
(a) 991026927
(b) 991026973
(c) 991036973
(d) 991227973
Show Answer & Explanation

Answer: (b) 991026973

Explanation:
1. (1000 − 3)³ = 10⁹ − 9(10⁶) + 27000 − 27 = 991026973.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 23: The product (4x + 3y)(16x² − 12xy + 9y²) equals
(a) 64x³ − 27y³
(b) 16x³ + 9y³
(c) 64x³ + 27y³
(d) 4x³ + 3y³
Show Answer & Explanation

Answer: (c) 64x³ + 27y³

Explanation:
1. It has the form (a + b)(a² − ab + b²) = a³ + b³ with a = 4x and b = 3y.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 24: The value of 28³ − 78³ + 50³ is
(a) 327600
(b) 0
(c) −327600
(d) −163800
Show Answer & Explanation

Answer: (c) −327600

Explanation:
1. 28 + (−78) + 50 = 0, so the sum = 3(28)(−78)(50) = −327600.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 25: If a + b + c = 9 and ab + bc + ca = 26, then a³ + b³ + c³ − 3abc equals
(a) 27
(b) 81
(c) 9
(d) 243
Show Answer & Explanation

Answer: (a) 27

Explanation:
1. a² + b² + c² = 81 − 52 = 29. Then a³ + b³ + c³ − 3abc = 9(29 − 26) = 27.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 26: The product (2x − 3)(2x + 5) equals
(a) 4x² + 4x − 15
(b) 4x² − 4x − 15
(c) 4x² + 2x − 15
(d) 2x² + 4x − 15
Show Answer & Explanation

Answer: (a) 4x² + 4x − 15

Explanation:
1. With y = 2x: (y − 3)(y + 5) = y² + 2y − 15 = 4x² + 4x − 15.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 27: The value of 103 × 96 is
(a) 9988
(b) 9788
(c) 9898
(d) 9888
Show Answer & Explanation

Answer: (d) 9888

Explanation:
1. (100 + 3)(100 − 4) = 10000 − 100 − 12 = 9888.

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 28: Factorise x³ − 6x² + 32.
(a) (x − 2)(x + 4)²
(b) (x + 2)(x + 4)²
(c) (x + 2)(x − 4)²
(d) (x + 2)(x² − 6x + 16)
Show Answer & Explanation

Answer: (c) (x + 2)(x − 4)²

Explanation:
1. Check x = −2: −8 − 24 + 32 = 0, so (x + 2) is a factor. Dividing: x³ − 6x² + 32 = (x + 2)(x² − 8x + 16) = (x + 2)(x − 4)².

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

Question 29: Factorise x⁴ + x²y² + y⁴.
(a) (x² + y²)²
(b) (x² + xy + y²)²
(c) (x² + xy + y²)(x² − xy + y²)
(d) (x² − y²)(x² + y²)
Show Answer & Explanation

Answer: (c) (x² + xy + y²)(x² − xy + y²)

Explanation:
1. x⁴ + 2x²y² + y⁴ − x²y² = (x² + y²)² − (xy)² = (x² + y² + xy)(x² + y² − xy).

Teacher's Note:
1. Recall the standard identities and apply them to the given expression before substituting numbers.
2. Always verify by expanding your answer or by substituting a simple value.

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