Chapter 2: Polynomials
Exercise & In-Text Questions & Answers

Answer & Reference:
(i) 4x² – 3x + 7 → Polynomial in one variable (only whole number exponents)
(ii) y² + √2 → Polynomial in one variable (only whole number exponents)
(iii) 3√t + t√2 → Not a polynomial (exponent of t is 1/2, not a whole number)
(iv) y + 2/y → Not a polynomial (can be written as y + 2y⁻¹, exponent -1 is not a whole number)
(v) x¹⁰ + y³ + t⁵⁰ → Not a polynomial in one variable (has three variables x, y, t)
(Reference: Page 26-27, Section 2.2)

Answer & Reference:
(i) 2 + x² + x → Coefficient of x² is 1
(ii) 2 – x² + x³ → Coefficient of x² is -1
(iii) (π/2)x² + x → Coefficient of x² is π/2
(iv) √2x – 1 → Coefficient of x² is 0 (since x² term is absent)
(Reference: Page 26, Section 2.2)

Answer & Reference:
Binomial of degree 35: x³⁵ + 5
Monomial of degree 100: 7x¹⁰⁰
(Reference: Page 27, Section 2.2)

Answer & Reference:
(i) 5x³ + 4x² + 7x → Degree is 3 (highest power of x)
(ii) 4 – y² → Degree is 2 (highest power of y)
(iii) 5t – 7 → Degree is 1 (highest power of t)
(iv) 3 → Degree is 0 (constant polynomial)
(Reference: Page 27, Example 1)

Answer & Reference:
(i) x² + x → Quadratic (degree 2)
(ii) x – x³ → Cubic (degree 3)
(iii) y + y² + 4 → Quadratic (degree 2)
(iv) 1 + x → Linear (degree 1)
(v) 3t → Linear (degree 1)
(vi) r² → Quadratic (degree 2)
(vii) 7x³ → Cubic (degree 3)
(Reference: Page 28, Section 2.2)

Answer & Reference:
Let p(x) = 5x – 4x² + 3
(i) p(0) = 5(0) – 4(0)² + 3 = 3
(ii) p(–1) = 5(–1) – 4(–1)² + 3 = –5 – 4 + 3 = –6
(iii) p(2) = 5(2) – 4(2)² + 3 = 10 – 16 + 3 = –3
(Reference: Page 29-30, Section 2.3)

Answer & Reference:
(i) p(y) = y² – y + 1
    p(0) = 0 – 0 + 1 = 1
    p(1) = 1 – 1 + 1 = 1
    p(2) = 4 – 2 + 1 = 3
(ii) p(t) = 2 + t + 2t² – t³
    p(0) = 2 + 0 + 0 – 0 = 2
    p(1) = 2 + 1 + 2 – 1 = 4
    p(2) = 2 + 2 + 8 – 8 = 4
(iii) p(x) = x³
    p(0) = 0
    p(1) = 1
    p(2) = 8
(iv) p(x) = (x – 1)(x + 1) = x² – 1
    p(0) = 0 – 1 = –1
    p(1) = 1 – 1 = 0
    p(2) = 4 – 1 = 3
(Reference: Page 31, Exercise 2.2)

Answer & Reference:
(i) p(x) = 3x + 1, x = –1/3 → p(–1/3) = 3(–1/3) + 1 = –1 + 1 = 0 → Yes
(ii) p(x) = 5x – π, x = 4/5 → p(4/5) = 5(4/5) – π = 4 – π ≠ 0 → No
(iii) p(x) = x² – 1, x = 1, –1 → p(1) = 1 – 1 = 0, p(–1) = 1 – 1 = 0 → Both are zeroes
(iv) p(x) = (x + 1)(x – 2), x = –1, 2 → p(–1) = 0, p(2) = 0 → Both are zeroes
(v) p(x) = x², x = 0 → p(0) = 0 → Yes
(vi) p(x) = lx + m, x = –m/l → p(–m/l) = l(–m/l) + m = –m + m = 0 → Yes
(vii) p(x) = 3x² – 1, x = –1/√3, 2/√3 → p(–1/√3) = 3(1/3) – 1 = 0, p(2/√3) = 3(4/3) – 1 = 3 ≠ 0 → Only –1/√3 is a zero
(viii) p(x) = 2x + 1, x = 1/2 → p(1/2) = 2(1/2) + 1 = 2 ≠ 0 → No
(Reference: Page 32, Exercise 2.2)

Answer & Reference:
(i) p(x) = x + 5 → x + 5 = 0 → x = –5
(ii) p(x) = x – 5 → x – 5 = 0 → x = 5
(iii) p(x) = 2x + 5 → 2x + 5 = 0 → x = –5/2
(iv) p(x) = 3x – 2 → 3x – 2 = 0 → x = 2/3
(v) p(x) = 3x → 3x = 0 → x = 0
(vi) p(x) = ax, a ≠ 0 → ax = 0 → x = 0
(vii) p(x) = cx + d, c ≠ 0 → cx + d = 0 → x = –d/c
(Reference: Page 32, Exercise 2.2)

Answer & Reference:
Using Factor Theorem: If p(–1) = 0, then (x+1) is a factor.
(i) p(x) = x³ + x² + x + 1 → p(–1) = –1 + 1 – 1 + 1 = 0 → Yes
(ii) p(x) = x⁴ + x³ + x² + x + 1 → p(–1) = 1 – 1 + 1 – 1 + 1 = 1 ≠ 0 → No
(iii) p(x) = x⁴ + 3x³ + 3x² + x + 1 → p(–1) = 1 – 3 + 3 – 1 + 1 = 1 ≠ 0 → No
(iv) p(x) = x³ – x² – (√2 + 1)x + √2 → p(–1) = –1 – 1 + (√2 + 1) + √2 = 2√2 – 1 ≠ 0 → No
(Reference: Page 32, Example 6 and Exercise 2.3)

Answer & Reference:
(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1
    Zero of g(x) is –1. p(–1) = –2 + 1 + 2 – 1 = 0 → Yes
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
    Zero of g(x) is –2. p(–2) = –8 + 12 – 6 + 1 = –1 ≠ 0 → No
(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3
    Zero of g(x) is 3. p(3) = 27 – 36 + 3 + 6 = 0 → Yes
(Reference: Page 32, Factor Theorem)

Answer & Reference:
Since x – 1 is a factor, p(1) = 0.
(i) p(x) = x² + x + k → 1 + 1 + k = 0 → k = –2
(ii) p(x) = 2x² + kx + √2 → 2 + k + √2 = 0 → k = –2 – √2
(iii) p(x) = kx² – √2x + 1 → k – √2 + 1 = 0 → k = √2 – 1
(iv) p(x) = kx² – 3x + k → k – 3 + k = 0 → 2k = 3 → k = 3/2
(Reference: Page 33, Example 7)

Answer & Reference:
(i) 12x² – 7x + 1 = 12x² – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1) = (3x – 1)(4x – 1)
(ii) 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)
(iii) 6x² + 5x – 6 = 6x² + 9x – 4x – 6 = 3x(2x + 3) – 2(2x + 3) = (2x + 3)(3x – 2)
(iv) 3x² – x – 4 = 3x² + 3x – 4x – 4 = 3x(x + 1) – 4(x + 1) = (x + 1)(3x – 4)
(Reference: Page 33-34, Example 8)

Answer & Reference:
(i) x³ – 2x² – x + 2 = x²(x – 2) –1(x – 2) = (x – 2)(x² – 1) = (x – 2)(x – 1)(x + 1)
(ii) x³ – 3x² – 9x – 5 = x³ + x² – 4x² – 4x – 5x – 5 = x²(x + 1) – 4x(x + 1) – 5(x + 1) = (x + 1)(x² – 4x – 5) = (x + 1)(x – 5)(x + 1) = (x + 1)²(x – 5)
(iii) x³ + 13x² + 32x + 20 = x³ + x² + 12x² + 12x + 20x + 20 = x²(x + 1) + 12x(x + 1) + 20(x + 1) = (x + 1)(x² + 12x + 20) = (x + 1)(x + 10)(x + 2)
(iv) 2y³ + y² – 2y – 1 = y²(2y + 1) –1(2y + 1) = (2y + 1)(y² – 1) = (2y + 1)(y – 1)(y + 1)
(Reference: Page 35, Exercise 2.3)

Answer & Reference:
Using Identity IV: (x + a)(x + b) = x² + (a + b)x + ab
(i) (x + 4)(x + 10) = x² + (4 + 10)x + 4×10 = x² + 14x + 40
(ii) (x + 8)(x – 10) = x² + (8 – 10)x + 8×(–10) = x² – 2x – 80
(iii) (3x + 4)(3x – 5) = (3x)² + (4 – 5)(3x) + 4×(–5) = 9x² – 3x – 20
(iv) (y² + 3/2)(y² – 3/2) = (y²)² – (3/2)² = y⁴ – 9/4 (Using Identity III: x² – y² = (x+y)(x–y))
(v) (3 – 2x)(3 + 2x) = 3² – (2x)² = 9 – 4x² (Using Identity III)
(Reference: Page 36, Example 11 and Identity IV)

Answer & Reference:
(i) 103 × 107 = (100 + 3)(100 + 7) = 100² + (3+7)100 + 3×7 = 10000 + 1000 + 21 = 11021
(ii) 95 × 96 = (100 – 5)(100 – 4) = 100² + (–5–4)100 + (–5)(–4) = 10000 – 900 + 20 = 9120
(iii) 104 × 96 = (100 + 4)(100 – 4) = 100² – 4² = 10000 – 16 = 9984
(Reference: Page 37, Example 12)

Answer & Reference:
(i) 9x² + 6xy + y² = (3x)² + 2(3x)(y) + y² = (3x + y)² (Using Identity I)
(ii) 4y² – 4y + 1 = (2y)² – 2(2y)(1) + 1² = (2y – 1)² (Using Identity II)
(iii) x² – y²/100 = x² – (y/10)² = (x – y/10)(x + y/10) (Using Identity III)
(Reference: Page 37, Example 13)

Answer & Reference:
Using Identity V: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
(i) (x + 2y + 4z)² = x² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x) = x² + 4y² + 16z² + 4xy + 16yz + 8zx
(ii) (2x – y + z)² = (2x)² + (–y)² + z² + 2(2x)(–y) + 2(–y)(z) + 2(z)(2x) = 4x² + y² + z² – 4xy – 2yz + 4zx
(iii) (–2x + 3y + 2z)² = (–2x)² + (3y)² + (2z)² + 2(–2x)(3y) + 2(3y)(2z) + 2(2z)(–2x) = 4x² + 9y² + 4z² – 12xy + 12yz – 8zx
(iv) (3a – 7b – c)² = (3a)² + (–7b)² + (–c)² + 2(3a)(–7b) + 2(–7b)(–c) + 2(–c)(3a) = 9a² + 49b² + c² – 42ab + 14bc – 6ac
(v) (–2x + 5y – 3z)² = (–2x)² + (5y)² + (–3z)² + 2(–2x)(5y) + 2(5y)(–3z) + 2(–3z)(–2x) = 4x² + 25y² + 9z² – 20xy – 30yz + 12zx
(vi) [1/4 a – 1/2 b + 1]² = (1/4 a)² + (–1/2 b)² + 1² + 2(1/4 a)(–1/2 b) + 2(–1/2 b)(1) + 2(1)(1/4 a) = 1/16 a² + 1/4 b² + 1 – 1/4 ab – b + 1/2 a
(Reference: Page 38, Example 14 and 15)

Answer & Reference:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz = (2x)² + (3y)² + (–4z)² + 2(2x)(3y) + 2(3y)(–4z) + 2(–4z)(2x) = (2x + 3y – 4z)²
(ii) 2x² + y² + 8z² – 2√2 xy + 4√2 yz – 8xz = (√2 x)² + y² + (2√2 z)² + 2(√2 x)(–y) + 2(–y)(2√2 z) + 2(2√2 z)(√2 x) = (√2 x – y + 2√2 z)²
(Reference: Page 38, Example 16)

Answer & Reference:
Using Identity VI: (x + y)³ = x³ + y³ + 3xy(x + y)
Using Identity VII: (x – y)³ = x³ – y³ – 3xy(x – y)
(i) (2x + 1)³ = (2x)³ + 1³ + 3(2x)(1)(2x + 1) = 8x³ + 1 + 12x² + 6x = 8x³ + 12x² + 6x + 1
(ii) (2a – 3b)³ = (2a)³ – (3b)³ – 3(2a)(3b)(2a – 3b) = 8a³ – 27b³ – 36a²b + 54ab²
(iii) (3/2 x + 1)³ = (3/2 x)³ + 1³ + 3(3/2 x)(1)(3/2 x + 1) = 27/8 x³ + 1 + 27/4 x² + 9/2 x
(iv) (x – 2/3 y)³ = x³ – (2/3 y)³ – 3(x)(2/3 y)(x – 2/3 y) = x³ – 8/27 y³ – 2x²y + 4/3 xy²
(Reference: Page 39, Example 17)

Answer & Reference:
(i) (99)³ = (100 – 1)³ = 100³ – 1³ – 3(100)(1)(100 – 1) = 1000000 – 1 – 29700 = 970299
(ii) (102)³ = (100 + 2)³ = 100³ + 2³ + 3(100)(2)(100 + 2) = 1000000 + 8 + 61200 = 1061208
(iii) (998)³ = (1000 – 2)³ = 1000³ – 2³ – 3(1000)(2)(1000 – 2) = 1000000000 – 8 – 5988000 = 994011992
(Reference: Page 39, Example 18)

Answer & Reference:
(i) 8a³ + b³ + 12a²b + 6ab² = (2a)³ + b³ + 3(2a)²b + 3(2a)b² = (2a + b)³
(ii) 8a³ – b³ – 12a²b + 6ab² = (2a)³ – b³ – 3(2a)²b + 3(2a)b² = (2a – b)³
(iii) 27 – 125a³ – 135a + 225a² = 3³ – (5a)³ – 3(3)²(5a) + 3(3)(5a)² = (3 – 5a)³
(iv) 64a³ – 27b³ – 144a²b + 108ab² = (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)² = (4a – 3b)³
(v) 27p³ – 1/216 – 9/2 p² + 1/4 p = (3p)³ – (1/6)³ – 3(3p)²(1/6) + 3(3p)(1/6)² = (3p – 1/6)³
(Reference: Page 40, Example 19)

Answer & Reference:
(i) RHS = (x + y)(x² – xy + y²) = x(x² – xy + y²) + y(x² – xy + y²) = x³ – x²y + xy² + x²y – xy² + y³ = x³ + y³ = LHS
(ii) RHS = (x – y)(x² + xy + y²) = x(x² + xy + y²) – y(x² + xy + y²) = x³ + x²y + xy² – x²y – xy² – y³ = x³ – y³ = LHS
(Reference: Page 40, Identity VIII derivation)

Answer & Reference:
Using identities from Question 23:
(i) 27y³ + 125z³ = (3y)³ + (5z)³ = (3y + 5z)((3y)² – (3y)(5z) + (5z)²) = (3y + 5z)(9y² – 15yz + 25z²)
(ii) 64m³ – 343n³ = (4m)³ – (7n)³ = (4m – 7n)((4m)² + (4m)(7n) + (7n)²) = (4m – 7n)(16m² + 28mn + 49n²)
(Reference: Page 41, Question 10)

Answer & Reference:
Using Identity VIII: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
27x³ + y³ + z³ – 9xyz = (3x)³ + y³ + z³ – 3(3x)(y)(z) = (3x + y + z)((3x)² + y² + z² – (3x)y – yz – (3x)z) = (3x + y + z)(9x² + y² + z² – 3xy – yz – 3xz)
(Reference: Page 40, Example 20)

Answer & Reference:
RHS = 1/2(x + y + z)[(x² – 2xy + y²) + (y² – 2yz + z²) + (z² – 2zx + x²)]
= 1/2(x + y + z)[2x² + 2y² + 2z² – 2xy – 2yz – 2zx]
= (x + y + z)(x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz (from Identity VIII)
= LHS
(Reference: Page 41, Question 12)

Answer & Reference:
From Identity VIII: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
Given x + y + z = 0, so:
x³ + y³ + z³ – 3xyz = 0 × (x² + y² + z² – xy – yz – zx) = 0
Therefore, x³ + y³ + z³ = 3xyz
(Reference: Page 41, Question 13)

Answer & Reference:
Using the result from Question 27: If x + y + z = 0, then x³ + y³ + z³ = 3xyz
(i) (–12)³ + (7)³ + (5)³: Since –12 + 7 + 5 = 0, value = 3(–12)(7)(5) = –1260
(ii) (28)³ + (–15)³ + (–13)³: Since 28 – 15 – 13 = 0, value = 3(28)(–15)(–13) = 16380
(Reference: Page 41, Question 14)

Answer & Reference:
(i) Area: 25a² – 35a + 12 = 25a² – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
    Possible length and breadth: (5a – 4) and (5a – 3)
(ii) Area: 35y² + 13y – 12 = 35y² + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3)
    Possible length and breadth: (5y + 4) and (7y – 3)
(Reference: Page 41, Question 15)

Answer & Reference:
(i) Volume: 3x² – 12x = 3x(x – 4)
    Possible dimensions: 3, x, (x – 4)
(ii) Volume: 12ky² + 8ky – 20k = 4k(3y² + 2y – 5) = 4k(3y² + 5y – 3y – 5) = 4k[y(3y + 5) – 1(3y + 5)] = 4k(3y + 5)(y – 1)
    Possible dimensions: 4k, (3y + 5), (y – 1)
(Reference: Page 42, Question 16)

Answer & Reference:

Concept	Description	Reference
Polynomial	An algebraic expression with whole number exponents of variables	Page 26
Degree of Polynomial	Highest power of the variable in the polynomial	Page 27
Linear Polynomial	Polynomial of degree 1: ax + b, a ≠ 0	Page 28
Quadratic Polynomial	Polynomial of degree 2: ax² + bx + c, a ≠ 0	Page 28
Cubic Polynomial	Polynomial of degree 3: ax³ + bx² + cx + d, a ≠ 0	Page 28
Zero of Polynomial	A number 'c' such that p(c) = 0	Page 30
Factor Theorem	x – a is a factor of p(x) iff p(a) = 0	Page 32
Algebraic Identities	Key identities for factorization and expansion	Page 36-40

(Reference: Page 42, Section 2.6 Summary)