CBSE Class 8 Mathematics Algebraic Expressions and Factorisation MCQs Set 14

Answer: (b) \( -7xy^2 \)

Explanation:
1. The term is (−7xy²) × z³, so the coefficient of z³ is the rest of the term.
2. Coefficient of z³ = \( -7xy^2 \).

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: (a) \( 2xy,\ -3xy \)

Explanation:
1. Like terms have exactly the same variables with the same powers.
2. 2xy and −3xy both contain x¹y¹, so they are like terms. In the other pairs the powers or variables differ.

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) \( 4x^2+xy+2y^2 \)

Explanation:
1. Collect like terms: x²: 2 + 5 − 3 = 4; y²: 3 − 1 = 2; xy: −5 + 6 = 1.
2. The simplified expression is \( 4x^2+xy+2y^2 \).

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: (d) \( x^2+x+1 \)

Explanation:
1. A trinomial has exactly three unlike terms.
2. x² + x + 1 has three unlike terms. The first two options simplify to binomials and xyz is a monomial.

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) less than 3

Explanation:
1. In division of polynomials the degree of the remainder is always less than the degree of the divisor.
2. The divisor has degree 3, so the remainder has degree less than 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: (b) \( 3t^2+11 \)

Explanation:
1. Since t⁶ = (t³ + 1)(t³ − 1) + 1, we have t⁶ + 3t² + 10 = (t³ + 1)(t³ − 1) + 3t² + 11.
2. The remainder is \( 3t^2+11 \) (its degree 2 is less than 3).

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: (d) \( -21x^3y^2z^3 \)

Explanation:
1. Multiply the coefficients: 3 × (−7) = −21.
2. Add the powers of each variable: x¹⁺² = x³, y¹⁺¹ = y², z³.
3. Product = \( -21x^3y^2z^3 \).

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: (a) 3

Explanation:
1. Divide each term by 2x: \( \frac x2+4x^2+1+3x^3 \).
2. The highest power of x is 3, so the degree is 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: (b) \( x^8-x^5+1 \)

Explanation:
1. A polynomial in standard form has its terms written in descending order of the powers of the variable (with whole-number powers).
2. Only \( x^8-x^5+1 \) is in descending order.

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) \( 12a^2b^2 \)

Explanation:
1. Difference = (a⁴ + 4a²b² + b⁴) − (a⁴ − 8a²b² + b⁴).
2. = 4a²b² + 8a²b² = \( 12a^2b^2 \).

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: (b) 3

Explanation:
1. Dividing by 2x gives 4x³ − (3/2)x + (5/2)x², whose highest power is 3.
2. The degree is three.

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) \( -5xy \)

Explanation:
1. −5x²y³ = (−5xy) × (xy²).
2. The coefficient of xy² is \( -5xy \).

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) factor

Explanation:
1. If the remainder is zero then dividend = quotient × divisor.
2. So the divisor is a factor of the dividend.

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) The degree of \( \frac{8x^2-3x}{2x} \) is one

Explanation:
1. 5x² − x³ + x² simplifies to 6x² − x³ (a binomial). x + x⁻¹ has a negative power, so it is not a polynomial. The degree of a polynomial is never negative.
2. (8x² − 3x)/(2x) = 4x − 3/2, whose degree is one, so the last statement is true.

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) 7

Explanation:
1. Square both sides: \( x^2+2+\frac1{x^2}=9 \).
2. So \( x^2+\frac1{x^2}=7 \).

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) 9

Explanation:
1. Square: \( x^2-2+\frac1{x^2}=\frac14 \), so \( x^2+\frac1{x^2}=\frac94 \).
2. Then \( 4\left(x^2+\frac1{x^2}\right)=9 \).

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) 194

Explanation:
1. \( x^2+\frac1{x^2}=16-2=14 \).
2. \( x^4+\frac1{x^4}=14^2-2=194 \).

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) 0

Explanation:
1. \( x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right) \).
2. = (√3)³ − 3√3 = 3√3 − 3√3 = 0.

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.