CBSE Class 8 Mathematics Algebraic Expressions and Factorisation MCQs Set 03

Answer: (d) −1

Explanation:
1. Put x = 2: 4 + 6q − 2q = 0, so 4q = −4 and q = −1.

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (c) −8

Explanation:
1. Put x = −2: −8 + 24 − 8 + k = 0, so k = −8.

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (c) x + 3

Explanation:
1. f(1) = 0, f(2) = 8 − 24 + 22 − 6 = 0 and f(3) = 27 − 54 + 33 − 6 = 0, so f(x) = (x − 1)(x − 2)(x − 3).
2. f(−3) ≠ 0, so (x + 3) is not a factor.

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (b) x − 2

Explanation:
1. Put x = 2: 16 − 40 + 20 − 20 + 24 = 0, so (x − 2) is a factor.
2. Put x = −2 and x = −4: neither gives zero.

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (d) 1, 2, 3

Explanation:
1. Statement 1: p(−3) = −27 + 18 − 9 + 18 = 0, true. Statement 2: p(−2) = −8 + 8 + 2 − 2 = 0, true. Statement 3: p(−1) = −1 + 1 + 4 − 4 = 0, true.
2. Statement 4: p(2) = 16 − 6 + 4 = 14 ≠ 0, false.
3. So statements 1, 2 and 3 are correct.

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (b) (x − 1) but not by (x + 1)

Explanation:
1. Put x = 1: 1 − 1 + 1 − 1 = 0, so (x − 1) is a factor.
2. Put x = −1: −1 + 1 − 1 − 1 = −2 ≠ 0, so (x + 1) is not a factor.

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (b) 1

Explanation:
1. Put x = 1: 1 − k = 0, so k = 1.

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (b) \( (2x-3y)(4x^2+9y^2+6xy) \)

Explanation:
1. Use a³ − b³ = (a − b)(a² + ab + b²) with a = 2x, b = 3y.
2. = (2x − 3y)(4x² + 6xy + 9y²).

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (b) \( (x+y)(x^2+y^2-xy+2x-2y) \)

Explanation:
1. x³ + y³ = (x + y)(x² − xy + y²) and 2x² − 2y² = 2(x + y)(x − y).
2. Taking (x + y) common: (x + y)(x² − xy + y² + 2x − 2y).

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (c) \( (x-2)^2(x-1) \)

Explanation:
1. Put x = 1: 1 − 5 + 8 − 4 = 0, so (x − 1) is a factor.
2. Dividing gives x² − 4x + 4 = (x − 2)², so the factors are (x − 1)(x − 2)².

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (c) \( (x^2+2x+2)(x^2-2x+2) \)

Explanation:
1. x⁴ + 4 = (x²)² + 2·x²·2 + 4 − 4x² = (x² + 2)² − (2x)².
2. = (x² + 2x + 2)(x² − 2x + 2).

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (a) \( 2y(3x^2+y^2) \)

Explanation:
1. (x + y)³ − (x − y)³ = (x³ + 3x²y + 3xy² + y³) − (x³ − 3x²y + 3xy² − y³) = 6x²y + 2y³.
2. = 2y(3x² + y²).

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (a) \( (x+2y)(x^2-2xy+4y^2) \)

Explanation:
1. a³ + b³ = (a + b)(a² − ab + b²) with a = x, b = 2y.
2. = (x + 2y)(x² − 2xy + 4y²).

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (c) m = 7, n = −18

Explanation:
1. Put x = 1: 1 + 10 + m + n = 0, so m + n = −11.
2. Put x = −2: −8 + 40 − 2m + n = 0, so −2m + n = −32.
3. Subtracting: 3m = 21, so m = 7 and n = −18.

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (b) 12

Explanation:
1. Remainder = p(−1) = −1 + 6 + 7 = 12.

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (a) 184

Explanation:
1. Remainder = p(3) = 243 − 81 + 36 − 14 = 184.

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (b) −3

Explanation:
1. Remainder = p(2) = 16 − 24 + 8 − 10 + 7 = −3.

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (d) −63

Explanation:
1. Remainder = p(−3) = −135 + 45 + 18 + 9 = −63.

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.