CBSE Class 8 Mathematics Algebraic Expressions and Factorisation MCQs Set 05

Answer: (b) \( \left(- x + 3 y + 1\right) \left(x - 3 y + 1\right) \)

Explanation:
1. 1 - (x² - 6xy + 9y²) = 1 - (x - 3y)².
2. Multiplying the factors back gives \( - x^{2} + 6 x y - 9 y^{2} + 1 \), 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: (b) \( \left(2 a + b\right) \left(- 2 x + 3 y\right) \)

Explanation:
1. Take (2a + b) common: (2a + b)[(x + y) - (3x - 2y)] = (2a + b)(-2x + 3y).
2. Multiplying the factors back gives \( - 4 a x + 6 a y - 2 b x + 3 b y \), 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: (a) \( \left(x - 2 y\right) \left(x - 2 y - 4\right) \)

Explanation:
1. -4x + 8y = -4(x - 2y), so take (x - 2y) common.
2. Multiplying the factors back gives \( x^{2} - 4 x y - 4 x + 4 y^{2} + 8 y \), 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) \( x^{2} \left(a - 2 b\right) \left(x + 1\right) \)

Explanation:
1. Take x²(a - 2b) common.
2. Multiplying the factors back gives \( a x^{3} + a x^{2} - 2 b x^{3} - 2 b x^{2} \), 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: (c) \( 15 x^{2} y^{2} \left(x + y\right) \)

Explanation:
1. The HCF of the two terms is 15x²y².
2. Multiplying the factors back gives \( 15 x^{3} y^{2} + 15 x^{2} y^{3} \), 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: (d) \( 3 a \left(3 - 4 a\right) \left(6 a - 7 b\right) \)

Explanation:
1. Take 3a(6a - 7b) common: 3a(6a - 7b)(3 - 4a).
2. Multiplying the factors back gives \( - 72 a^{3} + 84 a^{2} b + 54 a^{2} - 63 a b \), 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: (a) \( \left(x + 8 \sqrt{3}\right) \left(x - 2 \sqrt{3}\right) \)

Explanation:
1. 8√3 × (-2√3) = -48 and 8√3 - 2√3 = 6√3.
2. Multiplying the factors back gives \( x^{2} + 6 \sqrt{3} x - 48 \), 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) \( \left(x - 2\right) \left(3 x - 4\right) \)

Explanation:
1. 3 × 8 = 24; split -10 = -4 - 6: 3x² - 4x - 6x + 8 = x(3x - 4) - 2(3x - 4).
2. Multiplying the factors back gives \( 3 x^{2} - 10 x + 8 \), 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) \( \left(x + 5 y\right) \left(x - 3 y\right) \)

Explanation:
1. 5 × (-3) = -15 and 5 - 3 = 2.
2. Multiplying the factors back gives \( x^{2} + 2 x y - 15 y^{2} \), 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: (c) \( \left(x + 3\right) \left(x + 10\right) \)

Explanation:
1. Numbers with product 30 and sum 13: 10 and 3.
2. Multiplying the factors back gives \( x^{2} + 13 x + 30 \), 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) \( \left(x - 6\right) \left(x - 7\right) \)

Explanation:
1. Numbers with product 42 and sum -13: -7 and -6.
2. Multiplying the factors back gives \( x^{2} - 13 x + 42 \), 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) \( \left(x + 2\right) \left(x - 3\right) \)

Explanation:
1. Numbers with product -6 and sum -1: -3 and 2.
2. Multiplying the factors back gives \( x^{2} - x - 6 \), 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) \( \left(a + 1\right) \left(a - 1\right) \left(a^{2} + 1\right) \)

Explanation:
1. a⁴ - 1 = (a² + 1)(a² - 1) and a² - 1 = (a + 1)(a - 1), so the complete factorisation has three factors.
2. Multiplying the factors back gives \( a^{4} - 1 \), 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: (c) \( \left(a^{2} - b^{2}\right) \left(a^{4} + a^{2} b^{2} + b^{4}\right) \)

Explanation:
1. Use x³ - y³ with x = a², y = b²; (a² - b²)(a⁴ + a²b² + b⁴) is the result.
2. Multiplying the factors back gives \( a^{6} - b^{6} \), 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: (a) \( \left(- 3 m n + r^{2}\right)^{2} \)

Explanation:
1. This is (a - b)² with a = r² and b = 3mn.
2. Multiplying the factors back gives \( 9 m^{2} n^{2} - 6 m n r^{2} + r^{4} \), 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: (a) \( \left(x + 12\right) \left(x - 3\right) \)

Explanation:
1. Numbers with product -36 and sum 9: 12 and -3.
2. Multiplying the factors back gives \( x^{2} + 9 x - 36 \), 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) \( 2 \left(a + 6 b\right) \left(a - 6 b\right) \)

Explanation:
1. Take 2 common: 2(a² - 36b²) = 2(a - 6b)(a + 6b).
2. Multiplying the factors back gives \( 2 a^{2} - 72 b^{2} \), 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: (d) \( \left(- a + x\right) \left(x + 5\right) \)

Explanation:
1. Group: x(x - a) + 5(x - a) = (x + 5)(x - a).
2. Multiplying the factors back gives \( - a x - 5 a + x^{2} + 5 x \), 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: (c) \( n \left(n + 1\right) \left(n + 2\right) \)

Explanation:
1. Take n common: n(n² + 3n + 2) = n(n + 1)(n + 2).
2. Multiplying the factors back gives \( n^{3} + 3 n^{2} + 2 n \), 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) \( \left(2 x + 1\right) \left(4 x + 3\right) \)

Explanation:
1. 8 × 3 = 24 and 4 + 6 = 10: 8x² + 4x + 6x + 3 = 4x(2x + 1) + 3(2x + 1).
2. Multiplying the factors back gives \( 8 x^{2} + 10 x + 3 \), 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.