CBSE Class 8 Mathematics Algebraic Expressions and Factorisation MCQs Set 04

Answer: (c) \( \left(x y + 1\right) \left(x y + z\right) \)

Explanation:
1. Group as xy(xy + 1) + z(xy + 1) and take (xy + 1) common.
2. Multiplying the factors back gives \( x^{2} y^{2} + x y z + x y + z \), the given expression.

Teacher's Note:
1. Look for a common bracket and take it out as a factor.
2. Group terms in pairs when no single common factor exists.

Answer: (b) \( \left(3 b + c + 5 y\right) \left(3 b + c - 5 y\right) \)

Explanation:
1. Use a² - b² with a = c + 3b and b = 5y.
2. Multiplying the factors back gives \( 9 b^{2} + 6 b c + c^{2} - 25 y^{2} \), the given expression.

Teacher's Note:
1. A difference of two squares factorises as a² - b² = (a + b)(a - b).
2. Rewrite each term as a perfect square before applying the identity.

Answer: (a) \( \left(2 x + 3\right) \left(3 x + 4\right) \)

Explanation:
1. 6 × 12 = 72; split 17 = 8 + 9 so 6x² + 8x + 9x + 12 = 2x(3x + 4) + 3(3x + 4).
2. Multiplying the factors back gives \( 6 x^{2} + 17 x + 12 \), the given expression.

Teacher's Note:
1. To factorise x² + px + q find two numbers with sum p and product q.
2. Split the middle term and take out common factors in pairs.

Answer: (d) \( \left(- b + c + 1\right) \left(b - c + 1\right) \)

Explanation:
1. Use 1² - (b - c)² = (1 - (b - c))(1 + (b - c)).
2. Multiplying the factors back gives \( - b^{2} + 2 b c - c^{2} + 1 \), the given expression.

Teacher's Note:
1. A difference of two squares factorises as a² - b² = (a + b)(a - b).
2. Rewrite each term as a perfect square before applying the identity.

Answer: (b) \( 5 x^{3} \left(- 3 a^{3} - 2 a^{2} + x^{2}\right) \)

Explanation:
1. The HCF of the three terms is 5x³.
2. Multiplying the factors back gives \( - 15 a^{3} x^{3} - 10 a^{2} x^{3} + 5 x^{5} \), 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.

Answer: (a) \( \left(x y + 3\right) \left(x y - 8\right) \)

Explanation:
1. Treat xy as one variable: numbers with sum -5 and product -24 are 3 and -8.
2. Multiplying the factors back gives \( x^{2} y^{2} - 5 x y - 24 \), the given expression.

Teacher's Note:
1. To factorise x² + px + q find two numbers with sum p and product q.
2. Split the middle term and take out common factors in pairs.

Answer: (c) \( \left(a - 2 b\right) \left(a - 27 b\right) \)

Explanation:
1. Find two numbers with product 54 and sum -29: -2 and -27.
2. Multiplying the factors back gives \( a^{2} - 29 a b + 54 b^{2} \), the given expression.

Teacher's Note:
1. To factorise x² + px + q find two numbers with sum p and product q.
2. Split the middle term and take out common factors in pairs.

Answer: (b) \( \left(a - 16\right)^{2} \)

Explanation:
1. 256 = 16² and 32 = 2 × 16, so the trinomial is (a - 16)².
2. Multiplying the factors back gives \( a^{2} - 32 a + 256 \), the given expression.

Teacher's Note:
1. A perfect square trinomial is a² ± 2ab + b² = (a ± b)².
2. Check that the middle term equals twice the product of the roots of the outer terms.

Answer: (d) \( \left(21 a^{2} + x^{2}\right) \left(- 22 a^{2} + x^{2}\right) \)

Explanation:
1. Treat x² as one variable: numbers with product -462 and sum -1 are 21 and -22.
2. Multiplying the factors back gives \( - 462 a^{4} - a^{2} x^{2} + x^{4} \), the given expression.

Teacher's Note:
1. To factorise x² + px + q find two numbers with sum p and product q.
2. Split the middle term and take out common factors in pairs.

Answer: (c) \( \left(12 - x\right) \left(x + 17\right) \)

Explanation:
1. -x² - 5x + 204 = -(x² + 5x - 204) and 17 × 12 = 204 with 17 - 12 = 5, giving (17 + x)(12 - x).
2. Multiplying the factors back gives \( - x^{2} - 5 x + 204 \), the given expression.

Teacher's Note:
1. To factorise x² + px + q find two numbers with sum p and product q.
2. Split the middle term and take out common factors in pairs.

Answer: (a) \( \left(4 b + x^{2}\right) \left(- 4 b + x^{2}\right) \)

Explanation:
1. x⁴ = (x²)² and 16b² = (4b)².
2. Multiplying the factors back gives \( - 16 b^{2} + x^{4} \), the given expression.

Teacher's Note:
1. A difference of two squares factorises as a² - b² = (a + b)(a - b).
2. Rewrite each term as a perfect square before applying the identity.

Answer: (c) \( \left(a b + 2 c d\right) \left(a b - 2 c d\right) \)

Explanation:
1. Write as (ab)² - (2cd)².
2. Multiplying the factors back gives \( a^{2} b^{2} - 4 c^{2} d^{2} \), the given expression.

Teacher's Note:
1. A difference of two squares factorises as a² - b² = (a + b)(a - b).
2. Rewrite each term as a perfect square before applying the identity.

Answer: (a) \( \left(5 b + 9 p^{2} z^{3}\right) \left(- 5 b + 9 p^{2} z^{3}\right) \)

Explanation:
1. 81p⁴z⁶ = (9p²z³)² and 25b² = (5b)².
2. Multiplying the factors back gives \( - 25 b^{2} + 81 p^{4} z^{6} \), the given expression.

Teacher's Note:
1. A difference of two squares factorises as a² - b² = (a + b)(a - b).
2. Rewrite each term as a perfect square before applying the identity.

Answer: (d) \( \left(7 x + y + 1\right) \left(7 x + y - 1\right) \)

Explanation:
1. Use a² - 1² with a = 7x + y.
2. Multiplying the factors back gives \( 49 x^{2} + 14 x y + y^{2} - 1 \), the given expression.

Teacher's Note:
1. A difference of two squares factorises as a² - b² = (a + b)(a - b).
2. Rewrite each term as a perfect square before applying the identity.

Answer: (d) \( y \left(2 x + y - 16\right) \)

Explanation:
1. [(x + y - 8) - (x - 8)][(x + y - 8) + (x - 8)] = y(2x + y - 16).
2. Multiplying the factors back gives \( 2 x y + y^{2} - 16 y \), the given expression.

Teacher's Note:
1. A difference of two squares factorises as a² - b² = (a + b)(a - b).
2. Rewrite each term as a perfect square before applying the identity.

Answer: (b) \( y \left(2 x - y\right) \)

Explanation:
1. [x - (y - x)][x + (y - x)] = (2x - y)(y).
2. Multiplying the factors back gives \( 2 x y - y^{2} \), the given expression.

Teacher's Note:
1. A difference of two squares factorises as a² - b² = (a + b)(a - b).
2. Rewrite each term as a perfect square before applying the identity.

Answer: (a) \( \left(x^{2} + 2 x y + 4 y^{2}\right) \left(x^{2} - 2 x y + 4 y^{2}\right) \)

Explanation:
1. Add and subtract 4x²y²: (x² + 4y²)² - (2xy)² = (x² + 2xy + 4y²)(x² - 2xy + 4y²).
2. Multiplying the factors back gives \( x^{4} + 4 x^{2} y^{2} + 16 y^{4} \), the given expression.

Teacher's Note:
1. A perfect square trinomial is a² ± 2ab + b² = (a ± b)².
2. Check that the middle term equals twice the product of the roots of the outer terms.

Answer: (b) \( \left(- 4 y^{2} + z\right) \left(16 y^{4} + 4 y^{2} z + z^{2}\right) \)

Explanation:
1. z³ - (4y²)³ = (z - 4y²)(z² + 4zy² + 16y⁴).
2. Multiplying the factors back gives \( - 64 y^{6} + z^{3} \), the given expression.

Teacher's Note:
1. Use a³ - b³ = (a - b)(a² + ab + b²) and a³ + b³ = (a + b)(a² - ab + b²).
2. Write each term as a perfect cube first.

Answer: (c) \( \left(2 x - 9 y^{2}\right) \left(4 x^{2} + 18 x y^{2} + 81 y^{4}\right) \)

Explanation:
1. (2x)³ - (9y²)³ = (2x - 9y²)(4x² + 18xy² + 81y⁴).
2. Multiplying the factors back gives \( 8 x^{3} - 729 y^{6} \), the given expression.

Teacher's Note:
1. Use a³ - b³ = (a - b)(a² + ab + b²) and a³ + b³ = (a + b)(a² - ab + b²).
2. Write each term as a perfect cube first.

Answer: (b) \( \left(y^{2} + 5\right) \left(y^{2} - 7\right) \)

Explanation:
1. Treat y² as one variable: 5 and -7 have product -35 and sum -2.
2. Multiplying the factors back gives \( y^{4} - 2 y^{2} - 35 \), the given expression.

Teacher's Note:
1. To factorise x² + px + q find two numbers with sum p and product q.
2. Split the middle term and take out common factors in pairs.