CBSE Class 8 Mathematics Algebraic Expressions and Factorisation MCQs Set 06

Answer: (c) \( \left(x + 1\right) \left(x + 2\right) \left(x + 3\right) \)

Explanation:
1. x = -1, -2, -3 each make the polynomial zero, so (x + 1), (x + 2), (x + 3) are factors.
2. Multiplying the factors back gives \( x^{3} + 6 x^{2} + 11 x + 6 \), 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: (c) \( \left(x + 1\right) \left(x + 2\right) \left(x - 1\right) \)

Explanation:
1. Group: x²(x + 2) - 1(x + 2) = (x + 2)(x² - 1) = (x + 2)(x - 1)(x + 1).
2. Multiplying the factors back gives \( x^{3} + 2 x^{2} - x - 2 \), 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) \( \left(x + 1\right) \left(x - 2\right) \left(x - 5\right) \)

Explanation:
1. f(-1) = 0, f(2) = 0, f(5) = 0, so the factors are (x + 1), (x - 2), (x - 5).
2. Multiplying the factors back gives \( x^{3} - 6 x^{2} + 3 x + 10 \), 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) \( \left(x + 1\right) \left(x - 1\right) \left(x - 2\right) \left(x - 5\right) \)

Explanation:
1. Trial roots -1, 1, 2, 5 all make the polynomial zero.
2. Multiplying the factors back gives \( x^{4} - 7 x^{3} + 9 x^{2} + 7 x - 10 \), 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) \( \left(x + 1\right) \left(x + 2\right) \left(x - 2\right) \left(x - 3\right) \)

Explanation:
1. Trial roots -1, -2, 2 and 3 all give zero.
2. Multiplying the factors back gives \( x^{4} - 2 x^{3} - 7 x^{2} + 8 x + 12 \), 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) \( \left(x + 1\right) \left(x + 2\right) \left(x + 3\right) \left(x + 4\right) \)

Explanation:
1. All coefficients positive so all roots are negative: -1, -2, -3, -4 give zero.
2. Multiplying the factors back gives \( x^{4} + 10 x^{3} + 35 x^{2} + 50 x + 24 \), 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: (d) \( \left(x + 3\right) \left(x - 1\right) \left(x - 3\right) \left(2 x - 5\right) \)

Explanation:
1. Roots 1, 3, -3 and 5/2 make the polynomial zero.
2. Multiplying the factors back gives \( 2 x^{4} - 7 x^{3} - 13 x^{2} + 63 x - 45 \), 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) \( \left(- a - 5 b\right) \left(19 a^{2} + 55 a b + 43 b^{2}\right) \)

Explanation:
1. Use x³ - y³ with x = 2a + b and y = 3(a + 2b); x - y = -(a + 5b).
2. Multiplying the factors back gives \( - 19 a^{3} - 150 a^{2} b - 318 a b^{2} - 215 b^{3} \), 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) \( x y z \left(x^{2} y^{2} z^{2} + x^{2} y + x z + y z\right) \)

Explanation:
1. The HCF of the four terms is xyz.
2. Multiplying the factors back gives \( x^{3} y^{3} z^{3} + x^{3} y^{2} z + x^{2} y z^{2} + x y^{2} z^{2} \), 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: (b) 4

Explanation:
1. The coefficients 4, 20 and 40 have HCF 4.
2. The constant term 40 has no x, so x is not common.

Teacher's Note:
1. The HCF takes the common numerical factor and the lowest power of each common variable.
2. 4x² + 20x + 40 = 4(x² + 5x + 10).

Answer: (d) 14100

Explanation:
1. Take 47 common: 47(198 + 102) = 47 × 300.
2. 47 × 300 = 14100.

Teacher's Note:
1. Factorisation can make arithmetic easier.
2. Check: 198 × 47 = 9306 and 102 × 47 = 4794, sum 14100.

Answer: (c) 1

Explanation:
1. Numerator is a³ - b³ with a = 155, b = 55; denominator is a² + ab + b², so the fraction equals a - b = 100.
2. 100 ÷ 100 = 1.

Teacher's Note:
1. Use a³ - b³ = (a - b)(a² + ab + b²).
2. Recognise the identity instead of multiplying out.

Answer: (a) 27

Explanation:
1. Square both sides: x² - 2 + 1/x² = 25.
2. x² + 1/x² = 27.

Teacher's Note:
1. (a - b)² = a² - 2ab + b².
2. Here ab = x × 1/x = 1.

Answer: (b) 15

Explanation:
1. The numerator is a² - b² = (a + b)(a - b) with a = 12.4 and b = 2.6.
2. So the fraction equals a + b = 15.

Teacher's Note:
1. Cancel the common factor (a - b).
2. 12.4 + 2.6 = 15.

Answer: (d) 31

Explanation:
1. (a - b)² = a² + b² - 2ab, so a² + b² = 25 + 2 × 3.
2. = 31.

Teacher's Note:
1. Rearrange the identity (a - b)² = a² - 2ab + b².
2. Do not forget the 2 in 2ab.

Answer: (b) \( 3abc \)

Explanation:
1. a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).
2. With a + b + c = 0 the right side is 0, so a³ + b³ + c³ = 3abc.

Teacher's Note:
1. Use the identity for the sum of three cubes.
2. Check with a = 1, b = 1, c = -2: 1 + 1 - 8 = -6 = 3(1)(1)(-2).

Answer: (d) \( 3ab \)

Explanation:
1. The terms are 3a²b², 6ab² and 9a³b.
2. HCF of coefficients is 3; lowest power of a is a, of b is b.
3. HCF = 3ab.

Teacher's Note:
1. The HCF takes the numeric HCF and the lowest power of each variable.
2. 3ab(ab + 2b + 3a²) is the factorised form.

Answer: (b) \( x^6-7x^3-8 \)

Explanation:
1. (x - 2)(x² + 2x + 4) = x³ - 8 and (x + 1)(x² - x + 1) = x³ + 1.
2. (x³ - 8)(x³ + 1) = x⁶ + x³ - 8x³ - 8 = x⁶ - 7x³ - 8.

Teacher's Note:
1. Pair the factors to use a³ ± b³.
2. Multiply (x³ - 8)(x³ + 1) term by term.

Answer: (c) 0

Explanation:
1. By the remainder theorem put x = -1: -1 + 3 - 3 + 1 = 0.
2. The remainder is 0 (indeed the expression is (x + 1)³).

Teacher's Note:
1. x + 1 = 0 gives x = -1.
2. The polynomial equals (x + 1)³.

Answer: (a) 2

Explanation:
1. R1 = f(-1) = -1 + 2 + 5a - 7 = 5a - 6.
2. R2 = g(2) = 8 + 4a - 24 + 6 = 4a - 10.
3. 2R1 + R2 = 14a - 22 = 6, so a = 2.

Teacher's Note:
1. Apply the remainder theorem to each polynomial.
2. Check: a = 2 gives R1 = 4 and R2 = -2, and 8 - 2 = 6.