CBSE Class 8 Mathematics Algebraic Expressions and Factorisation MCQs Set 07

Mathematics Objective Questions and Answers: Algebraic Expressions and Factorisation

Access targeted multiple-choice questions for Algebraic Expressions and Factorisation designed to align with the latest CBSE academic syllabus for Class 8 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Algebraic Expressions and Factorisation MCQs with Answers

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Multiple Choice Questions

Question 1: A restaurant has 8 square tables of side x m, 10 tables of size x m by 1 m and 3 tables of size 1 m by 1 m. The total area of the table tops in terms of x is:
(a) \( 18x+3 \)
(b) \( 8x^2+10x+3 \)
(c) \( 8x^2+10x+3x^2 \)
(d) \( 8x^2+13x \)
Show Answer & Explanation

Answer: (b) \( 8x^2+10x+3 \)

Explanation:
1. Area of square tables = 8 × x² = 8x².
2. Area of x by 1 tables = 10 × x = 10x; area of 1 by 1 tables = 3.
3. Total = 8x² + 10x + 3.

Teacher's Note:
1. Area of a rectangle = length × breadth.
2. Add the areas of the three kinds of tables.

Question 2: If all those tables (8x² + 10x + 3 m² in total) are joined to form one large rectangle, its dimensions are:
(a) \( (4x+3)\text{ m by }(2x+3)\text{ m} \)
(b) \( (4x+1)\text{ m by }(2x+3)\text{ m} \)
(c) \( (4x+3)\text{ m by }(2x+1)\text{ m} \)
(d) \( (8x+3)\text{ m by }(x+1)\text{ m} \)
Show Answer & Explanation

Answer: (c) \( (4x+3)\text{ m by }(2x+1)\text{ m} \)

Explanation:
1. Factorise the area: 8x² + 10x + 3 = 8x² + 4x + 6x + 3 = 4x(2x + 1) + 3(2x + 1).
2. = (4x + 3)(2x + 1), so the sides are (4x + 3) m and (2x + 1) m.

Teacher's Note:
1. Area of a rectangle = length × breadth, so factorise the area.
2. Check: (4x + 3)(2x + 1) = 8x² + 10x + 3.

Question 3: The first four terms of the sequence with nth term \( T_n=n^3+3n^2+2n \) are:
(a) 3, 12, 30, 60
(b) 6, 24, 60, 100
(c) 6, 24, 60, 120
(d) 6, 24, 54, 120
Show Answer & Explanation

Answer: (c) 6, 24, 60, 120

Explanation:
1. T₁ = 1 + 3 + 2 = 6; T₂ = 8 + 12 + 4 = 24.
2. T₃ = 27 + 27 + 6 = 60; T₄ = 64 + 48 + 8 = 120.

Teacher's Note:
1. Substitute n = 1, 2, 3, 4.
2. T_n = n(n + 1)(n + 2), the product of three consecutive integers.

Question 4: Every term of the sequence \( T_n=n^3+3n^2+2n \) is the product of three consecutive integers, so every term is always divisible by:
(a) 4
(b) 6
(c) 8
(d) 5
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. T_n = n(n + 1)(n + 2).
2. Among three consecutive integers one is a multiple of 3 and at least one is even, so the product is divisible by 6.

Teacher's Note:
1. Test with terms 6, 24, 60, 120: all are multiples of 6.
2. 24 is not a multiple of 5 and 6 is not a multiple of 4.

Question 5: Find the HCF of \( x^3 \) and \( x \).
(a) \( x^3 \)
(b) x
(c) \( x^2 \)
(d) 1
Show Answer & Explanation

Answer: (b) x

Explanation:
1. x³ = x × x × x and x = x.
2. The common factor with the lowest power is x.

Teacher's Note:
1. HCF takes the lowest power of the common variable.
2. x divides x³.

Question 6: Factorise \( 5x^2+5x \).
(a) \( x(5x+1) \)
(b) \( 5(x^2+x^2) \)
(c) \( 5x(x-1) \)
(d) \( 5x(x+1) \)
Show Answer & Explanation

Answer: (d) \( 5x(x+1) \)

Explanation:
1. The HCF of 5x² and 5x is 5x.
2. 5x² + 5x = 5x(x + 1).

Teacher's Note:
1. Divide each term by the HCF.
2. Check: 5x(x + 1) = 5x² + 5x.

Question 7: The common factors of \( 12x^3 \) and \( 6x \) are (HCF):
(a) \( 12x \)
(b) \( 2x \)
(c) \( 6x^3 \)
(d) \( 6x \)
Show Answer & Explanation

Answer: (d) \( 6x \)

Explanation:
1. HCF of 12 and 6 is 6; lowest power of x is x.
2. HCF = 6x.

Teacher's Note:
1. Numerical HCF times lowest variable power.
2. 6x divides both: 12x³ = 6x × 2x² and 6x = 6x × 1.

Question 8: The HCF of \( 6abc,\ 36ab^2c \) and \( 216a^2bc^2 \) is:
(a) \( 216abc \)
(b) \( 6a^2b^2c^2 \)
(c) \( 36abc \)
(d) \( 6abc \)
Show Answer & Explanation

Answer: (d) \( 6abc \)

Explanation:
1. HCF of 6, 36, 216 is 6.
2. Lowest powers of a, b and c are a, b and c.

Teacher's Note:
1. Compare the powers of each letter across the three terms.
2. 6abc divides all three terms.

Question 9: The HCF of \( 25x^2y^2,\ 15x^3 \) and \( 3x^4y \) is:
(a) \( x^2y \)
(b) \( x^2 \)
(c) \( 5x^2 \)
(d) \( 3x^2 \)
Show Answer & Explanation

Answer: (b) \( x^2 \)

Explanation:
1. HCF of 25, 15 and 3 is 1.
2. Lowest power of x is x²; y is not present in 15x³.
3. HCF = x².

Teacher's Note:
1. Letters absent from any term are not part of the HCF.
2. The numbers 25, 15 and 3 have no common factor greater than 1.

Question 10: Factorise \( 9x-81 \).
(a) \( 9(x-81) \)
(b) \( 9(x-9) \)
(c) \( 9x(1-9) \)
(d) \( 3(3x-27) \)
Show Answer & Explanation

Answer: (b) \( 9(x-9) \)

Explanation:
1. 9 is the HCF of 9 and 81.
2. 9x - 81 = 9(x - 9).

Teacher's Note:
1. Divide each term by 9.
2. 3(3x - 27) is not fully factorised.

Question 11: Factorise \( 7a-49b \).
(a) \( a(7-49b) \)
(b) \( 7(a-7b) \)
(c) \( 7(a-49b) \)
(d) \( 7(a+7b) \)
Show Answer & Explanation

Answer: (b) \( 7(a-7b) \)

Explanation:
1. HCF of 7 and 49 is 7.
2. 7a - 49b = 7(a - 7b).

Teacher's Note:
1. Divide both terms by 7.
2. 49b ÷ 7 = 7b.

Question 12: Factorise \( 8x^2+32x \).
(a) \( 8(x^2+4x) \)
(b) \( 8x(x+4) \)
(c) \( 8x(x+32) \)
(d) \( 4x(2x+8) \)
Show Answer & Explanation

Answer: (b) \( 8x(x+4) \)

Explanation:
1. HCF of 8x² and 32x is 8x.
2. 8x² + 32x = 8x(x + 4).

Teacher's Note:
1. HCF = numerical HCF × lowest power of x.
2. 8(x² + 4x) still has a common factor x.

Question 13: Factorise \( 25x^2y-10xy^2 \).
(a) \( 5xy(5x-2y) \)
(b) \( 5x(5xy-2y^2) \)
(c) \( xy(25x-10y) \)
(d) \( 5xy(5x+2y) \)
Show Answer & Explanation

Answer: (a) \( 5xy(5x-2y) \)

Explanation:
1. HCF of the terms is 5xy.
2. 25x²y ÷ 5xy = 5x and 10xy² ÷ 5xy = 2y.

Teacher's Note:
1. Take out the complete HCF.
2. The sign between the terms stays minus.

Question 14: Factorise \( (x+1)^2-(y-1)^2 \).
(a) \( (x-y)(x+y+2) \)
(b) \( (x-y+2)(x+y) \)
(c) \( (x+y)(x-y) \)
(d) \( (x+y+2)^2 \)
Show Answer & Explanation

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

Explanation:
1. Use a² - b² with a = x + 1 and b = y - 1.
2. [(x + 1) - (y - 1)][(x + 1) + (y - 1)] = (x - y + 2)(x + y).

Teacher's Note:
1. Remember to subtract the whole bracket: -(y - 1) = -y + 1.
2. Check at x = 1, y = 1: 4 - 0 = 4 and (2)(2) = 4.

Question 15: Simplify \( \frac{16a^2-81b^2}{4a-9b} \).
(a) \( a+b \)
(b) \( 4a-9b \)
(c) \( 4a+9b \)
(d) \( 4a+81b \)
Show Answer & Explanation

Answer: (c) \( 4a+9b \)

Explanation:
1. 16a² - 81b² = (4a + 9b)(4a - 9b).
2. Cancel (4a - 9b) to get 4a + 9b.

Teacher's Note:
1. Recognise the difference of squares.
2. Cancel only after factorising.

Question 16: Is \( 4z-3 \) a factor of \( 4z^2-13z-12 \)?
(a) No, because the remainder is not zero
(b) No, because the degree is different
(c) Yes, because the quotient is z - 2
(d) Yes, because the remainder is 3
Show Answer & Explanation

Answer: (a) No, because the remainder is not zero

Explanation:
1. Put z = 3/4: 4(9/16) - 13(3/4) - 12 = 9/4 - 39/4 - 12 = -39/2.
2. The remainder -39/2 is not zero, so 4z - 3 is not a factor.

Teacher's Note:
1. By the factor theorem the divisor divides the polynomial only if the value at its root is zero.
2. Long division gives quotient z - 5/2 and remainder -39/2.

Question 17: Divide \( 2y^4+7y^2+8y^3+4y+3 \) by \( y+3 \). The quotient is:
(a) \( 2y^3+2y^2+y+1 \)
(b) \( 2y^3+2y^2+y-1 \)
(c) \( 2y^3-2y^2+y+1 \)
(d) \( 2y^3+y^2+2y+1 \)
Show Answer & Explanation

Answer: (a) \( 2y^3+2y^2+y+1 \)

Explanation:
1. Arrange in descending powers: 2y⁴ + 8y³ + 7y² + 4y + 3.
2. Dividing by y + 3 gives 2y³ + 2y² + y + 1 with remainder 0.
3. Check: (y + 3)(2y³ + 2y² + y + 1) = 2y⁴ + 8y³ + 7y² + 4y + 3.

Teacher's Note:
1. Always write the dividend in descending order of powers first.
2. Verify by multiplying quotient by divisor.

Question 18: Factorise \( 6 x y^{2} - 5 x y - 6 x \).
(a) \( x \left(2 y + 3\right) \left(3 y - 2\right) \)
(b) \( \left(2 y - 3\right) \left(3 y + 2\right) \)
(c) \( x \left(2 y - 3\right) \left(3 y + 2\right) \)
(d) \( x \left(2 y + 2\right) \left(3 y - 3\right) \)
Show Answer & Explanation

Answer: (c) \( x \left(2 y - 3\right) \left(3 y + 2\right) \)

Explanation:
1. Take x common: x(6y² - 5y - 6) and split -5y = -9y + 4y.
2. Multiplying the factors back gives \( 6 x y^{2} - 5 x y - 6 x \), the given expression.

Teacher's Note:
1. Take out the HCF of all the terms first and write what is left inside the bracket.
2. Check by multiplying the factors back.

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Multiple Choice Questions (MCQs) for Class 8 Mathematics Algebraic Expressions and Factorisation

Class 8 Mathematics Algebraic Expressions and Factorisation Objective Test Questions

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FAQs

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