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Polynomials-NCERT Solutions
∵ All the exponents of y are whole numbers. ∴ 
is a polynomial in one variable. 
(v) x10 + y3 + t50 ∵; Exponent of every variable is a whole number, ∴ x10 + y3 + t50 is a polynomial in x, y and t, i.e. in three variables. 2. Write the co-efficients of x2 in each of the following: (i) 2 + x2 + x (ii) 2 – x2 + x3 (iii) 
(v) 
Ans. (i) 2 + x2 + x The co-efficient of x2 is 1. (ii) 2 – x2 + x3 The co-efficient of x2 is (–1). (iii) 
The co-efficient of x2 is 
(iv) 
∴ The co-efficient of x2 is 0 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100. Ans. (i) A binomial of degree 35 can be: 3x35 – 4 (ii) A monomial of degree 100 can be: 
4. Write the degree of each of the following polynomials: (i) 5x3 + 4x2 + 7x (ii) 4 - y2 (iii) 
(iv) 3 Ans. (i) 5x3 + 4x2 + 7x ∵ The highest exponent of x is 3. ∴ The degree of the polynomial is 3. (ii) 4 – y2 ∵ The highest exponent of y is 2. ∴ The degree of the polynomial is 2. (iii) ∵ The highest exponent of t is 1. ∴ The degree of the polynomial is 1. (iv) 3 since, 3 = 3x° ∴ The degree of the polynomial 3 is 0. 5. Classify the following as linear, quadratic and cubic polynomials: (i) x2 + x (ii) x – x3 (iii) y + y2 + 4 (iv) 1 + x (v) 3t (vi) r2 (vii) 7x3 Ans. (i) x2 +x ∵ The degree of x2 + x is 2. ∴ It is a quadratic polynomial. (ii) x – x3 ∵ The degree of x – x3 is 3. ∴ It is a cubic polynomial. (iii) y + y2 + 4 ∵ The degree of y + y2 + 4 is 2. ∴ It is a quadratic polynomial. (iv) 1 + x ∵ The degree of 1 + x is 1. ∴ It is a linear polynomial. (v) 3t ∵ The degree of 3t is 1. ∴ It is a linear polynomial. (vi) r2 ∵ The degree of r2 is 2. ∴ It is a quadratic polynomial. (vii) 7x3 ∵ The degree of 7x3 is 3. ∴ It is a cubic polynomial. Exercise– 2.2 1. Find the value of the polynomial 5x – 4x2 + 3 at (i) x = 0 (ii) x = –1 (iii) x = 2 Ans. (i) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(0) = 5(0) – 4(0) + 3 = 0 – 0 + 3 = 3 Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. (ii) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(–1) = 5(–1) – 4(–1)2 + 3 = – 5 – 4(1) + 3 = –5 – 4 + 3 = –9 + 3 = –6 ∴ The value of 5x – 4x2 + 3 at x = –1 is –6. (iii) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(2) = 5(2) – 4(2)2 + 3 = –10 – 4(4) + 3 = 10 – 16 + 3 = –3 Thus the value of 5x – 4x2 + 3 at x = 2 is –3 2. Find p(0), p(1) and p(2) for each of the following polynomials: (i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t2 – t3 (iii) p(x) = x3 (iv) p(x) = (x – 1) (x + 1) Ans. (i) p(y) = y2 – y + 1 ∵ p(y) = y2 – y + 1 = (y)2 – y + 1 ∴ p(0) = (0)2 – (0) + 1 = 0 – 0 + 1 = 1 p(1) = (1)2 – (1) + 1 = 1 – 1 + 1 = 1 p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3 (ii) p(t) = 2 + t + 2t2 – t3 ∵ p(t) = 2 + t + 2t2 – t3 = 2 + t + 2(t)2 – (t)3 ∴ p(0) = 2 + (0) + 2(0)2 – (0)3 = 2 + 0 + 0 – 0 = 2 p(1) = 2 + (1) + 2(1)2 – (1)3 = 2 + 1 + 2 – 1 = 4 p(2) = 2 + 2 + 2(2)2 – (2)3 = 2 + 2 + 8 – 8 = 4 (iii) p(x) = x3 ∵ p(x) = x3 = (x)3 ∴ p(0) = (0)3 = 0 p(1) = (1)3 = 1 p(2) = (2)3 = 8 [∵ 2 × 2 × 2 = 8] (iv) p(x)= (x – 1)(x + 1) ∵ p(x) = (x – 1)(x + 1) ∴ p(0) = (0 – 1)(0 + 1) = –1 × 1 = –1 p(1) = (1 – 1)(1 + 1) = (0)(2) = 0 p(2) = (2 – 1)(2 + 1) = (1)(2) = 3 3. Verify whether the following are zeros of the polynomial, indicated against them. 
Ans. (i) ∵ p(x) = 3x + 1 
(iii) Since, p(x) = x2 – 1 ∴ p(1) = (1)2 – 1 = 1 – 1 = 0 Since, p(1) = 0, ∴ x = 1 is a zero of x2 – 1. Also p(–1) = (–1)2 – 1 = 1 – 1 = 0 i.e. p(–1) = 0, ∴ x = –1 is also a zero of x2 – 1. (iv) We have p(x) = (x + 1)(x – 2) ∴ p(–1) = (–1 + 1)(–1 – 2) = (0)(–3) = 0 Since p(–1) = 0, ∴ x = –1 is a zero of (x + 1)(x – 1). ∴ Also, p(2) = (2 + 1)(2 – 2) = (3)(0) = 0 ∴ Since p(2) = 0, ∴ x = 2 is also a zero of (x + 1)(x – 1). (v) We have p(x) = x2 ∴ p(0) = (0)2 = 0 Since p(0) = 0, ∴ 0 is a zero of x2. (vi) We have p(x) = lx + m 
(vii) We have p(x) = 3x2 – 1 
(viii) We have p(x) = 2x + 1 
is not a zero of 2x + 1. 4. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. Ans. (i) We have p(x) = x + 5 ∴ p(x) = 0 ⇒ x + 5 = 0 ∴ or x = –5 ∴ Thus, a zero of x + 5 is (–5). (ii) We have p(x) = x – 5 ∴ p(x) = 0 ⇒ x – 5 = 0 ∴ or x = 5 ∴ Thus, a zero of x – 5 is 5. (iii) We have p(x) = 2x + 5 ∴ p(x) = 0 ⇒ 2x + 5 = 0 or 2x = –5 
Thus, a zero of 3x – 2 is 
(v) Since p(x) = 3x ∴ p(x) = 0 ⇒ 3x = 0 
Thus, a zero of 3x is 0. (vi) Since, p(x) = ax, a ≠ 0 ∴ p(x) = 0 ⇒ ax = 0 or 
Thus, a zero of ax is 0. (viii) Since, p(x) = cx + d ∴ p(x) = 0 ⇒ cx + d = 0 or cx = –d or 
Thus, a zero of cx + d is 
5. If p(x) = x2 – 4x + 3, evaluate: 
Ans. We have p(x) = x2 4x + 3 ∴ p(–1) = (–1)2 – 4(–1) + 3 = 1 + 4 + 3 = 8 and p(2) = (2)2 – 4(2) + 3 = 4 – 8 + 3 = –1 
Exercise– 2.3 1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by (i) x + 1 (ii) 
(iii) x (iv) x + π (v) 5 + 2x Ans. (i) ∴ The zero of x + 1 is –1 And by remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x + 1, then remainder is p(–1). ∴ p(–1) = (–1)3 + 3 (–1)2 + 3(–1) + 1 = –1 + (3 × 1) + (–3) + 1 = –1 + 3 – 3 + 1 = 0 Thus, the required = 0 
and p(x) = x3 + 3x2 + 3x + 1 ∴ For divisor 
remainder is given as 
(iii) We have p(x) = x3 + 3x2 + 3x + 1 and the zero of x is 0 p(0) = (0)3 + 3(0)2 + 3(0) + 1 = 0 + 0 + 0 +1 = 1 Thus, the required remainder = 1. (iv) We have p(x) = x3 + 3x2 + 3x + 1 and zero of x + p = (–p) [∵ x + π = 0 ⇒ x = –π] ∴ p(–π) = (–5)3 + 3(–π)2 + 3(–π) + 1 = –π3 + 3(π2) + (–3π) + 1 = –π3 + 3π2 – 3π + 1 Thus, the required remainder is –π3 + 3π2 – 3π + 1. (v) We have (p(x) = x3 + 3x2 + 3x + 1 and zero of 5 + 2x is 
2. Find the remainder when x3 – ax2 + 6x – a is divided by x – a. Ans. We have p(x) = x3 – ax2 + 6x – a ∵ Zero of x – a is a. [∵ x – a = 0 ⇒ x = a] ∴ p(a) = (a)3 a(a)2 + 6(a) – a = a3 – a3 + 6a – a = 0 + 5a = 5a ∴ Thus, the required remainder = 5a 3. Check whether 7 + 3x is a factor of 3x3 + 7x. Ans. We have p(x) = 3x3 + 7x and zero of 7 + 3x is 
i.e. the remainder is not 0. ∴ 3x3 – 7x is not divisible by 7 + 3x. Thus, (7 + 3x) is not a factor of 3x3 – 7x. Exercise– 2.4 1. Determine which of the following polynomials has a factor (x + 1): (i) x3 + x2 + x + 1 (ii) x4 + x + x2 + x + 1 (iii) x4 + 3x3 + 3x2 + x +1 (iv) 
Ans. For x + 1 = 0, we have x = –1. (i) p(x) = x3 + x2 + x + 1 ∴ p(–1) = (–1)3 + (–1)2 + (–1) + 1 = –1 + 1 – 1 + 1 = 0 i.e. when p(x) is divided by (x + 1), then the remainder is zero. ∴ (x + 1) is a factor of x3 + x2 + x + 1. (ii) p(x) = x4 + x3 + x2 +x + 1 ∴ p(–1) = (–1)4 + (–1)3 + (–1)2 + (–1) + 1 = 1 – 1 + 1 – 1 + 1 = 3 – 2 = 1 ∵ p(x) is not divisible by x + 1. i.e. (x + 1) is not a factor of x4 + x3 + x2 + x + 1. (iii) ∵ p(x) = x4 + x3 + x2 +x + 1 ∴ p(–1) = (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1 = (1) + 3(– 1) + 3(1) + (– 1) + 1 = 1 – 3 + 3 – 1 + 1 = 1 ≠ 0 ∵ f(–1) ≠ 0 ∴ (x + 1) is not a factor of x4 + 3x3 + 3x2 +x +1. 
2. Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1 (ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3 Ans. (i) We have p(x) = 2x3 + x2 – 2x – 1 and g(x) = x + 1 ∴ p(–1) = 2(–1)3 + (–1)2 – 2(–1) – 1 = 2(–1)3 + (–1)2 – 2(–1) – 1 = 2(–1) + 1 + 2 – 1 = –2 + 1 + 2 – 1 = –3 + 3 = 0 ∵ p(–1) = 0 ∴ g(x) is a factor of p(x). (ii) We have p(x) = x3 + 3x2 + 3x + 1 and g(x) = x + 2 ∴ p(–2) = (–2)3 + 3(–2)2 + 3(–2) + 1 = –8 + 3(4) + (–6) + 1 = –8 + 12 – 6 + 1 = –8 – 6 + 12 + 1 = –14 + 13 = –1 ∵ p(–2) ≠ 0 Thus, g(x) is not a facot of p(x). (iii) We have p(x) = x3 – 4x2 + x + 6 and g(x) = x – 3 ∴ p(3) = (3)3 – 4(3)2 + (3) + 6 = 27– 4(9) + 3 + 6 = 27– 36 + 3 + 6 = 0 Since g(x) = 0 ∴ g(x) is a factor of p(x). 3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: 
Ans. Here p(x) = x3 + x + k For x – 1 be a factor of p(x), p(1) should be equal to 0. We have p(1) = (1)3 + 1 + k or p(1) = 1 + 1 + k = k + 2 ∴ k + 2 = 0 ⇒ k = –2 (ii) Here, p(x) = 2x2 + kx + 2 For x – 1, be a factor of p(x), p(1) = 0 Since, 
∵ p(1) must be equal to 0. ⇒ k = –2 – 2 or k = –(2 + 2 ). (iii) 
∴ For (x – 1) be a factor of p(x), p(1) should be equal to 0. 
(iv) Here p(x) = kx2 – 3x + k and g(x) = x – 1 For g(x) be a factor of p(x), p(1) should be equal to 0. Since p(1) = k(1)2 – 3(1) + k = k – 3 + k = 2k – 3 ∴ 2k – 3 = 0 
4. Factorise: (i) 12x2 – 7x + 1 (ii) 2x2 + 7x + 3 (iii) 6x2 + 5x – 6 (iv) 3x2 – x – 4 Ans. (i) 12x2 – 7x + 1 Here co-efficient of x2 = 12 Co-efficient of x = –7 and constant term = 1 ∴ a = 12, b = –7, c = 1 Now, l + m = –7 and lm = ac = 12 × 1 ∴ We have l = (–4) and m = (–3) i.e. b = –7 + (–4 –3). Now, 12x2 –7x + 1 = 12x2 – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1) = (3x – 1)(3x – 1) Thus, 12x2 – 7x + 1 = (3x – 1)(3x – 1) (ii) 2x2 + 7x + 3 Here, a = 12, b = –7, c = 1 ∴ l + m = 7 and lm = 2 × 3 = 6 i.e. l + 6 = 7 and 1 × 6 = 6∴ l = 1 and m = 6 We have 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 1) Thus, 2x2 + 7x + 3 = (2x + 1)(x + 1) (iii) 6x2 + 5x – 6 We have a = 6, b = 5 and c = –6 ∴ l + m = 5 and lm = ac = 6 × (–6) = –36 ∴ l + m = 9 + (–4) ∴ 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = 3x(2x + 3) – 2(2x + 3) = (2x + 3)(3x – 2) Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2) (iv) 3x2 – x – 4 We have a = 3, b = –1 and c = –4 ∴ l + m = –1 and lm = 3 × (–4) = –12 ∴ l = – 4 and m = 3 Now, 3x2 – x – 4 = 3x2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1) Thus, 3x2 – x – 4 = (3x – 4)(x + 1) 5. Factorise: (i) x3 – 2x2 – x + 2 (ii) x3 – 3x2 – 9x – 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 – 2y – 1 Ans. (i) x3 – 2x2 – x + 2 Rearranging the terms, we have x3 – 2x2 – x + 2 = x3 – x – 2x2 + 2 = x(x2 – 1) – 2(x2 – 1) = (x2 – 1) (x – 2) = [(x2 – (1)2][x – 2] = (x – 1) (x + 1)(x – 2) [∴ a2 – b2 = (a + b)(a – b)] Thus, x3 – 2x2 – x + 2 = (x – 1) (x + 1)(x – 2) (ii) x3 – 3x2 – 9x – 5 We have p(x) = x3 – 3x2 – 9x – 5 By trial, let us find: p(1) = (1)3 – 3(1)2 – 9(1) –5 = 3 – 3 – 9 – 5 = –14 ≠ 0 Now p(–1) = (–1)3 – 3(–1)2 – 9(–1) –5 = –1 – 3(1) + 9 – 5 = –1 – 3 + 9 – 5 = 0 ∴ By factor theorem, [x – (–1)] is a factor of p(x). Now, 
∴ x3 – 3x2 – 9x – 5 = (x + 1)(x2 – 4x – 5) = (x + 1)[x2 – 5x + x – 5] [Splitting –4 into –5 and +1] = (x + 1)[x(x – 5) +1(x – 5] = (x + 1)[(x – 5) (x + 1)] = (x + 1)(x – 5)(x + 1) (iii) x3 + 13x2 + 32x + 20 We have p(x) = x3 + 13x2 + 32x + 20 By trial, let us find: p(1) = (1)3 + 13(1)2 + 32(1) + 20 = 1 + 13 + 32 + 20 = 66 ≠ 0 Now p(–1) = (–1)3 + 13(–1)2 + 32(–1) + 20 = –1 + 13 – 32 + 20 = 0 ∴ By factor theorem, [x – (–1)], i.e. (x + 1) is a factor p(x). 
or x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) = (x + 1) (x2 + 2x + 12x + 20) [Splitting the middle term] = (x + 1)[x(x + 2) + 10(x + 2)] = (x + 1)[(x + 2) (x + 10)] = (x + 1)(x + 2) (x + 10) (iv) 2y3 + y2 – 2y – 1 We have p(y) = 2y3 + y2 – 2y – 1 By trial, we have p(1) = 2(1)3 + (1)2 – 2(1) – 1 = 2(1) + 1 – 2 – 1 = 2 + 1 – 2 – 1 = 0 ∴ By factor theorem, (y – 1) is a factor of p(y). 
∴ 2y3 – y2 – 2y – 1 = (y – 1)(2y2 + 3y + 1) = (y – 1)[2y2 + 2y + y + 1) [Splitting the middle term] = (y – 1)[2y(y + 1) + 1(y + 1)] = (y – 1)[(y + 1) (2y + 1)] = (y – 1)(y + 1) (2y + 1) Exercise– 2.5 1. Use suitable identities to find the following products: (i) (x + 4) (x + 10) (ii) (x + 8) (x – 10) (iii) (3x + 4)(3x – 5) (iv) 
(v) (3 – 2x) (3 + 2x) Ans. (i) (x + 4) (x + 10): Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, we have: (x + 4)(x + 10) = x2 + (14 + 10)x + (4 × 10) = x2 + 14x + 40 (ii) (x + 8) (x – 10): Here, a = 8 and b = (–10) ∴ Using (x + a)(x + b) = x2 + (a + b) + ab, we have: (x + 8)(x – 10) = x2 + [8 + (–10)]x + [8 × (–10)] = x2 + [–2]x + [–80] = x2 + 2x – 80 (iii) (3x + 4)(3x – 5): ∴ Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, we have: (3x + 4)(3x – 5) = (3x)2 + [4 + (–5)]3x + [4 × (–5)] = 9x2 + [–1]3x + [–20] = 9x2 – 3x – 20 (iv) 
Using the identity (a + b)(a – b) = a2 – b2, we have: (v) (3 – 2x)(3 + 2x): Using the identity (a + b)(a – b) = a2 – b2, we have: (3 – 2x)(3 + 2x) = (3)2 – (2x)2 = 9 – 4x2 2. Evaluate the following products without multiplying directly: (i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96 Ans. (i) We have 103 × 107 = (100 + 3)(100 + 7) = (100)2 + (3 + 7) × 100 + (3 × 7) [Using (x + a)(x + b) = x2 + (a + b)x +ab] = 10000 + (10) × 100 + 21 = 10000 + 1000 + 21 = 11021 (ii) We have 95 × 96 = (100 – 5)(100 – 4) = (100)2 + [(–5) + (–4)] × 100 + [(–5) × (–4)] [Using (x + a)(x + b) = x2 + (a + b)x +ab] = 10000 + [–9] × 100 + 21 = 10000 + (–900) + 20 = 9120 (iii) We have 104 × 96 = (100 + 4)(100 – 4) = (100)2 – (4)2 [Using (a + b)(a – b) = a2 – b2] = 10000 – 16 = 9984 3. Factorise the following using appropriate identities: (i) 9x2 + 6xy + y2 (ii) 4y2 – 4y + 1 (iii) 
Ans. (i) We have 9x2 + 6xy + y2 = (3x)2 + 2(3x)(y) + (y)2 = (3x + y)2 [Using a2 + 2ab + b2 = (a + b)2] = (3x + y)(3x + y) (ii) We have 4y2 + 4y + 1 = (2y)2 – 2(2y)(1) + (1)2 = (2y – 1)2 [∴ a2 – 2ab + b2 = (a – b)2] = (2y – 1)(2y – 1) 
[Using a2 – b2 = (a + b)(a – b)] 4. Expand each of the following using suitable identities: (i) (x + 2y + 4z)2 (ii) (2x – y + z)2 (iii) (–2x + 3y + 2z)2 (iv) (3a – 7b – c)2 (v) (–2x + 5y – 3z)2 
Ans. (i) (x + 2y + 4z)2 We have (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx ∴ (x + 2y + 4z)2 = (x)2 + (2y)2 + (4y)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x) = x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx (ii) (xx – y + z)2 Using (x + y + z)2= x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (2x – y + z)2 = (2x)2 + (–y)2 + (z)2 + 2(2x)(–y) + 2(–y)(z) + 2(z)(2x) = 4x2 + y2 + z2 – 4xy – 2yz + 4zx (iii) (–2x + 3y + 2z)2 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (–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 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (3a – 7b – c)2 = (3a)2 + (–7b)2 + (–c)2 + 2(3a)(–7b) + 2(–7b)(–c) + 2(–c)(3a) = 9a2 + 49b2 + c2 + (–42ab) + (14bc) – 6ca = 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca (v) (–2x + 5y – 3z)2 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have ∴ (–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] = 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx 
Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have 
5. Factorise: 
Ans. (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz = (2x)2 + (3y)2 + (–4z)2 + 2(2x)(3y) + 2(3y)(–4z) + 2(–4z)(2x) [Using Identity V] = (2x + 3y – 4z)2 = (2x + 3y – 4z)(2x + 3y – 4z) 
6. Write the following cubes in expanded form: (i) (2x + 1)3 (ii) 2a – 3b)3 (iii) 
(iv) 
Ans. Using Identity VI and Identity VII, we have (x + y)3 = x3 + y3 + 3xy (x + y), and (x + y)3 = x3 + y3 + 3xy (x – y). (i) (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1) [(2x) + (1)] = 8x3 + 1 + 6x[2x + 1] [Using Identity VI] = 8x3 + 1 + 12x2 + 6x = 8x3 + 12x2 + 6x + 1 (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b) [(2a) + (3b)] = 8a3 – 27b3 – 18ab[2a – 3b] [Using Identity VII] = 8a3 – 27b3 – [36a2b – 54ab2] = 8a3 – 27b3 – 36a2b + 54ab2 
7. Evaluate the following using suitable identities: (i) (99)3 (ii) (102)3 (iii) (998)3 Ans. (i) (99)3 We have 99 = 100 – 1 ∴ 993 = (100 – 1)3 = 1003 – 13 – 3(100)(1)(100 – 1) = 1000000 – 1 – 30000 + 300 = 100300 – 30001 = 970299 (ii) (102)3 We have 102 = 100 + 2 (102)3 = (100 + 2)3 = (100)3 + (2)3 + 3(100)(2)[100 2 1] = 1000000 + 8 + 600[100 + 2] = 1000000 + 8 + 60000 + 1200 = 1061208 (ii) (998)3 We have 998 = 100 – 2 ∴ (999)3 = (1000 – 2)3 = (1000)3 – (2)3 – 3(1000)(2)[1000 – 2] = 10000000 – 8 – 6000[1000 – 2] = 10000000 – 8 – 600000 – 12000 = 994011992 8. Factorise each of the following: (i) 8a3 + b3 + 12a2b + 6ab2 (ii) 8a3 – b3 – 12a2b – 6ab2 (iii) 27 – 125a3 – 135a + 225a2 (iv) 64a3 – 27b3 – 144a2b + 108ab2 Ans. (i) 8a3 + b3 + 12a2b + 6ab2 = (2a)3 + (b)3 + 6ab(2a + b) = (2a)3 + (b)3 + 3(2a)(b)(2a + b) = (2a + b)3 [Using Identify VI] = (2a + b)(2a + b)(2a + b) (ii) 8a3 – b3 – 12a2b – 6ab2 = (2a)3 – (b)3 – 3(2a)(b)(2a – b) = (2a – b)3 [Using Identify VII] = (2a – b)(2a – b)(2a – b) (iii) 27 – 125a3 – 135a + 225a2 = (3)3 – (5a)3 – 3(3)(5a)[3 – 5a] = (3 – 5a)3 [Using Identify VII] = (3 – 5a)(3 – 5a)(3 – 5a) (iv) 64a3 – 27b3 – 144a2b + 108ab2 = (4a)3 – (3b)3 – 3(4a)(3b)[4a – 3b] = (4a – 3b)3 [Using Identify VII] = (4a – 3b)(4a – 3b)(4a – 3b) 
9. Verify: (i) x3 + y3 = (x + y)(x2 – xy + y2) (ii) x3 – y3 = (x – y)(x2 + xy + y2) Ans. (i) R.H.S. = (x + y)(x2 – xy + y2) = x(x2 – xy + y2) + y(x2 – xy + y2) = x3 – x2y + xy2 + x2y – xy2 + y3 = x3 + y3 = L.H.S (ii) R.H.S. = (x – y)(x2 + xy + y2) = x(x2 + xy + y2) – y(x2 + xy + y2) = x3 + x2y + xy2 – x2y – xy2 – y3 = x3 – y3 = L.H.S 10. Factorise each of the following: (i) 27y3 + 125z3 (ii) 64m3 – 343n3 Remember I. x3 + y3 = (x + y)(x2 + y2 – xy) II. x3 – y3 = (x – y)(x2 + y2 + xy) Ans. (i) Using the identity x3 + y3 = (x + y)(x2 + – xy + y2), we have 27y3 + 125z3 = (3y)3 + (5z)3 = (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2] = (3y + 5z)(9y2 – 15yz + 25z2) (ii) Using the identity x3 – y3 = (x – y)(x2 + xy + y2), we have 64m3 – 343n3 = (4m)3 – (7n)3 = (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2] = (4m + 7n)(16m2 + 28mn + 19n2) 11. Factorise 27x3 + y3 + z3 – 9xyz. Ans. Remember x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) We have 27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) ∴ Using the identity x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx), we have (3x)3 + y3 +z3 – 3(3x)(y)(z) = (3x + y +z)[(3x)2 + y2 + z2 – (3x × y) – (y × z) – (z × 3x)] = (3x + y +z)(9x2 + y2 + z2 – 3xy – yz – 3zx) 12. Verify that 
13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz. Ans. Ans. Since x + y + z = 0 ∴ x + y = –z or (x + y)3 = (–z)3 or x3 + y3 + 3xy(x + y) = –z3 or x3 + y3 + 3xy(–z) = –z3 [ x + y = (–z)] or x3 + y3 – 3xy = –z3 or (x3 + y3 + z3) – 3xy = 0 or (x3 + y3 + z3) = 3xy Hence, If x + y + z = 0, then (x3 + y3 + z3 = 3xy. 14. Without actually calculating the cubes, find the value of each of the following: (i) (–12)3 + (7)3 + (5)3 (ii) (28)3 + (–15)3 + (–13)3 Ans. (i) (–12)3 + (7)3 + (5)3 Let x = –12, y = 7 and z = 5 Then x + y + z = –12 + 7 + 5 = 0 We know that if x + y + z =0, then x3 + y3 + z3 = 3xyz. ∴ (–12)3 + (7)3 + (5)3 = 3[(–12)(7)(5)] [∴ (–12) + 7 + 5 = 0] = 3[–420] = –1260 Thus, (–12)3 + (7)3 + (5)3 = –1260 (ii) (28)3 + (–15)3 + (–13)3 Let x = 28, y = –15 and z = –13 ∴ x + y + z =0, then x3 + y3 + z3 = 3xyz. We know that if x + y + z =0, then x3 + y3 + z3 = 3xyz. (28)3 + (–15)3 + (–13)3 = 3(28)(–15)(–13) [28 + (–15) + (–13) = 0] = 3(5460) = 16380 Thus, (28)3 + (–15)3 + (–13)3 = 1638 15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: Area: 25a – 35a 2 + 12 Area: 35y2 + 13y – 12 (i) (ii) Remember Area of a rectangle = (Length) × (Breadth) Ans. (i) Area = 25a2 – 35a + 12 We have to factorise the polynomial: 25a2 – 35a + 12 Splitting the co-efficient of a, we have –35 = (–20) + (–15) [∴ 25 × 12 = 300 and (–20) × (–15) = 300] = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) Thus, the possible length and breadth are (5a –3) and (5a – 4). (ii) Area = 35y2 + 13y – 12 We have to factorise the polynomial: 35y2 + 13y – 12 Splitting the middle term, we get 13y = 28y – 15y [∴ 28 × (–15) = –420 and –12 × 35 = –420] ∴ 35y2 + 13y – 12 = 35y2 + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3) Thus, the possible length and breadth are (7y – 3) and (5y + 4). 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume: 3x – 122 x (ii) Volume: 12ky2 + 18ky – 12k Remember Volume of a cuboid = (Length) × (Breadth) × (Height) Ans. (i) Volume = 3x2 – 12x On factorising 3x2 – 12x, we have 3x2 – 12x = 3[x2 – 4x] = 3 × [x(x – 4)] = 3 × x × (x – 4) ∴The possible dimensions of the cuboid are: 3, x and (x – 4) units. (ii) Volume = 12ky2 + 8ky – 20k We have 12ky2 + 8ky – 20k = 4[3ky2 + 2ky – 5k] = 4[2(3y2 + 2y – 5] (Splitting the middle term) = 4k[3y(y – 1) + 5(y – 1)] = 4k[(3y + 5)(y – 1)] = 4k × (3y + 5) × (y – 1) Thus, the possible dimensions are: 4k, (3y + 5) and (y – 1) units.
Polynomials-NCERT Solutions
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∵ All the exponents of y are whole numbers. ∴ is a polynomial in one variable. (v) x10 + y3 + t50 ∵; Exponent of every variable is a whole number, ∴ x10 + y3 + t50 is a polynomial in x, y and t, i.e. in three variables. 2. Write the co-efficients of x2 in each of the following: (i) 2 + x2 + x (ii) 2 – x2 + x3 (iii) (v) Ans. (i) 2 + x2 + x The co-efficient of x2 is 1. (ii) 2 – x2 + x3 The co-efficient of x2 is (–1). (iii) The co-efficient of x2 is (iv) ∴ The co-efficient of x2 is 0 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100. Ans. (i) A binomial of degree 35 can be: 3x35 – 4 (ii) A monomial of degree 100 can be: 4. Write the degree of each of the following polynomials: (i) 5x3 + 4x2 + 7x (ii) 4 - y2 (iii) (iv) 3 Ans. (i) 5x3 + 4x2 + 7x ∵ The highest exponent of x is 3. ∴ The degree of the polynomial is 3. (ii) 4 – y2 ∵ The highest exponent of y is 2. ∴ The degree of the polynomial is 2. (iii) ∵ The highest exponent of t is 1. ∴ The degree of the polynomial is 1. (iv) 3 since, 3 = 3x° ∴ The degree of the polynomial 3 is 0. 5. Classify the following as linear, quadratic and cubic polynomials: (i) x2 + x (ii) x – x3 (iii) y + y2 + 4 (iv) 1 + x (v) 3t (vi) r2 (vii) 7x3 Ans. (i) x2 +x ∵ The degree of x2 + x is 2. ∴ It is a quadratic polynomial. (ii) x – x3 ∵ The degree of x – x3 is 3. ∴ It is a cubic polynomial. (iii) y + y2 + 4 ∵ The degree of y + y2 + 4 is 2. ∴ It is a quadratic polynomial. (iv) 1 + x ∵ The degree of 1 + x is 1. ∴ It is a linear polynomial. (v) 3t ∵ The degree of 3t is 1. ∴ It is a linear polynomial. (vi) r2 ∵ The degree of r2 is 2. ∴ It is a quadratic polynomial. (vii) 7x3 ∵ The degree of 7x3 is 3. ∴ It is a cubic polynomial. Exercise– 2.2 1. Find the value of the polynomial 5x – 4x2 + 3 at (i) x = 0 (ii) x = –1 (iii) x = 2 Ans. (i) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(0) = 5(0) – 4(0) + 3 = 0 – 0 + 3 = 3 Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. (ii) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(–1) = 5(–1) – 4(–1)2 + 3 = – 5 – 4(1) + 3 = –5 – 4 + 3 = –9 + 3 = –6 ∴ The value of 5x – 4x2 + 3 at x = –1 is –6. (iii) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(2) = 5(2) – 4(2)2 + 3 = –10 – 4(4) + 3 = 10 – 16 + 3 = –3 Thus the value of 5x – 4x2 + 3 at x = 2 is –3 2. Find p(0), p(1) and p(2) for each of the following polynomials: (i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t2 – t3 (iii) p(x) = x3 (iv) p(x) = (x – 1) (x + 1) Ans. (i) p(y) = y2 – y + 1 ∵ p(y) = y2 – y + 1 = (y)2 – y + 1 ∴ p(0) = (0)2 – (0) + 1 = 0 – 0 + 1 = 1 p(1) = (1)2 – (1) + 1 = 1 – 1 + 1 = 1 p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3 (ii) p(t) = 2 + t + 2t2 – t3 ∵ p(t) = 2 + t + 2t2 – t3 = 2 + t + 2(t)2 – (t)3 ∴ p(0) = 2 + (0) + 2(0)2 – (0)3 = 2 + 0 + 0 – 0 = 2 p(1) = 2 + (1) + 2(1)2 – (1)3 = 2 + 1 + 2 – 1 = 4 p(2) = 2 + 2 + 2(2)2 – (2)3 = 2 + 2 + 8 – 8 = 4 (iii) p(x) = x3 ∵ p(x) = x3 = (x)3 ∴ p(0) = (0)3 = 0 p(1) = (1)3 = 1 p(2) = (2)3 = 8 [∵ 2 × 2 × 2 = 8] (iv) p(x)= (x – 1)(x + 1) ∵ p(x) = (x – 1)(x + 1) ∴ p(0) = (0 – 1)(0 + 1) = –1 × 1 = –1 p(1) = (1 – 1)(1 + 1) = (0)(2) = 0 p(2) = (2 – 1)(2 + 1) = (1)(2) = 3 3. Verify whether the following are zeros of the polynomial, indicated against them. Ans. (i) ∵ p(x) = 3x + 1 (iii) Since, p(x) = x2 – 1 ∴ p(1) = (1)2 – 1 = 1 – 1 = 0 Since, p(1) = 0, ∴ x = 1 is a zero of x2 – 1. Also p(–1) = (–1)2 – 1 = 1 – 1 = 0 i.e. p(–1) = 0, ∴ x = –1 is also a zero of x2 – 1. (iv) We have p(x) = (x + 1)(x – 2) ∴ p(–1) = (–1 + 1)(–1 – 2) = (0)(–3) = 0 Since p(–1) = 0, ∴ x = –1 is a zero of (x + 1)(x – 1). ∴ Also, p(2) = (2 + 1)(2 – 2) = (3)(0) = 0 ∴ Since p(2) = 0, ∴ x = 2 is also a zero of (x + 1)(x – 1). (v) We have p(x) = x2 ∴ p(0) = (0)2 = 0 Since p(0) = 0, ∴ 0 is a zero of x2. (vi) We have p(x) = lx + m (vii) We have p(x) = 3x2 – 1 (viii) We have p(x) = 2x + 1 is not a zero of 2x + 1. 4. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. Ans. (i) We have p(x) = x + 5 ∴ p(x) = 0 ⇒ x + 5 = 0 ∴ or x = –5 ∴ Thus, a zero of x + 5 is (–5). (ii) We have p(x) = x – 5 ∴ p(x) = 0 ⇒ x – 5 = 0 ∴ or x = 5 ∴ Thus, a zero of x – 5 is 5. (iii) We have p(x) = 2x + 5 ∴ p(x) = 0 ⇒ 2x + 5 = 0 or 2x = –5 Thus, a zero of 3x – 2 is (v) Since p(x) = 3x ∴ p(x) = 0 ⇒ 3x = 0 Thus, a zero of 3x is 0. (vi) Since, p(x) = ax, a ≠ 0 ∴ p(x) = 0 ⇒ ax = 0 or Thus, a zero of ax is 0. (viii) Since, p(x) = cx + d ∴ p(x) = 0 ⇒ cx + d = 0 or cx = –d or Thus, a zero of cx + d is 5. If p(x) = x2 – 4x + 3, evaluate: Ans. We have p(x) = x2 4x + 3 ∴ p(–1) = (–1)2 – 4(–1) + 3 = 1 + 4 + 3 = 8 and p(2) = (2)2 – 4(2) + 3 = 4 – 8 + 3 = –1 Exercise– 2.3 1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by (i) x + 1 (ii) (iii) x (iv) x + π (v) 5 + 2x Ans. (i) ∴ The zero of x + 1 is –1 And by remainder theorem, when p(x) = x3 + 3x2 + 3x + 1 is divided by x + 1, then remainder is p(–1). ∴ p(–1) = (–1)3 + 3 (–1)2 + 3(–1) + 1 = –1 + (3 × 1) + (–3) + 1 = –1 + 3 – 3 + 1 = 0 Thus, the required = 0 and p(x) = x3 + 3x2 + 3x + 1 ∴ For divisor remainder is given as (iii) We have p(x) = x3 + 3x2 + 3x + 1 and the zero of x is 0 p(0) = (0)3 + 3(0)2 + 3(0) + 1 = 0 + 0 + 0 +1 = 1 Thus, the required remainder = 1. (iv) We have p(x) = x3 + 3x2 + 3x + 1 and zero of x + p = (–p) [∵ x + π = 0 ⇒ x = –π] ∴ p(–π) = (–5)3 + 3(–π)2 + 3(–π) + 1 = –π3 + 3(π2) + (–3π) + 1 = –π3 + 3π2 – 3π + 1 Thus, the required remainder is –π3 + 3π2 – 3π + 1. (v) We have (p(x) = x3 + 3x2 + 3x + 1 and zero of 5 + 2x is 2. Find the remainder when x3 – ax2 + 6x – a is divided by x – a. Ans. We have p(x) = x3 – ax2 + 6x – a ∵ Zero of x – a is a. [∵ x – a = 0 ⇒ x = a] ∴ p(a) = (a)3 a(a)2 + 6(a) – a = a3 – a3 + 6a – a = 0 + 5a = 5a ∴ Thus, the required remainder = 5a 3. Check whether 7 + 3x is a factor of 3x3 + 7x. Ans. We have p(x) = 3x3 + 7x and zero of 7 + 3x is i.e. the remainder is not 0. ∴ 3x3 – 7x is not divisible by 7 + 3x. Thus, (7 + 3x) is not a factor of 3x3 – 7x. Exercise– 2.4 1. Determine which of the following polynomials has a factor (x + 1): (i) x3 + x2 + x + 1 (ii) x4 + x + x2 + x + 1 (iii) x4 + 3x3 + 3x2 + x +1 (iv) Ans. For x + 1 = 0, we have x = –1. (i) p(x) = x3 + x2 + x + 1 ∴ p(–1) = (–1)3 + (–1)2 + (–1) + 1 = –1 + 1 – 1 + 1 = 0 i.e. when p(x) is divided by (x + 1), then the remainder is zero. ∴ (x + 1) is a factor of x3 + x2 + x + 1. (ii) p(x) = x4 + x3 + x2 +x + 1 ∴ p(–1) = (–1)4 + (–1)3 + (–1)2 + (–1) + 1 = 1 – 1 + 1 – 1 + 1 = 3 – 2 = 1 ∵ p(x) is not divisible by x + 1. i.e. (x + 1) is not a factor of x4 + x3 + x2 + x + 1. (iii) ∵ p(x) = x4 + x3 + x2 +x + 1 ∴ p(–1) = (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1 = (1) + 3(– 1) + 3(1) + (– 1) + 1 = 1 – 3 + 3 – 1 + 1 = 1 ≠ 0 ∵ f(–1) ≠ 0 ∴ (x + 1) is not a factor of x4 + 3x3 + 3x2 +x +1. 2. Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1 (ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3 Ans. (i) We have p(x) = 2x3 + x2 – 2x – 1 and g(x) = x + 1 ∴ p(–1) = 2(–1)3 + (–1)2 – 2(–1) – 1 = 2(–1)3 + (–1)2 – 2(–1) – 1 = 2(–1) + 1 + 2 – 1 = –2 + 1 + 2 – 1 = –3 + 3 = 0 ∵ p(–1) = 0 ∴ g(x) is a factor of p(x). (ii) We have p(x) = x3 + 3x2 + 3x + 1 and g(x) = x + 2 ∴ p(–2) = (–2)3 + 3(–2)2 + 3(–2) + 1 = –8 + 3(4) + (–6) + 1 = –8 + 12 – 6 + 1 = –8 – 6 + 12 + 1 = –14 + 13 = –1 ∵ p(–2) ≠ 0 Thus, g(x) is not a facot of p(x). (iii) We have p(x) = x3 – 4x2 + x + 6 and g(x) = x – 3 ∴ p(3) = (3)3 – 4(3)2 + (3) + 6 = 27– 4(9) + 3 + 6 = 27– 36 + 3 + 6 = 0 Since g(x) = 0 ∴ g(x) is a factor of p(x). 3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: Ans. Here p(x) = x3 + x + k For x – 1 be a factor of p(x), p(1) should be equal to 0. We have p(1) = (1)3 + 1 + k or p(1) = 1 + 1 + k = k + 2 ∴ k + 2 = 0 ⇒ k = –2 (ii) Here, p(x) = 2x2 + kx + 2 For x – 1, be a factor of p(x), p(1) = 0 Since, ∵ p(1) must be equal to 0. ⇒ k = –2 – 2 or k = –(2 + 2 ). (iii) ∴ For (x – 1) be a factor of p(x), p(1) should be equal to 0. (iv) Here p(x) = kx2 – 3x + k and g(x) = x – 1 For g(x) be a factor of p(x), p(1) should be equal to 0. Since p(1) = k(1)2 – 3(1) + k = k – 3 + k = 2k – 3 ∴ 2k – 3 = 0 4. Factorise: (i) 12x2 – 7x + 1 (ii) 2x2 + 7x + 3 (iii) 6x2 + 5x – 6 (iv) 3x2 – x – 4 Ans. (i) 12x2 – 7x + 1 Here co-efficient of x2 = 12 Co-efficient of x = –7 and constant term = 1 ∴ a = 12, b = –7, c = 1 Now, l + m = –7 and lm = ac = 12 × 1 ∴ We have l = (–4) and m = (–3) i.e. b = –7 + (–4 –3). Now, 12x2 –7x + 1 = 12x2 – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1) = (3x – 1)(3x – 1) Thus, 12x2 – 7x + 1 = (3x – 1)(3x – 1) (ii) 2x2 + 7x + 3 Here, a = 12, b = –7, c = 1 ∴ l + m = 7 and lm = 2 × 3 = 6 i.e. l + 6 = 7 and 1 × 6 = 6∴ l = 1 and m = 6 We have 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 1) Thus, 2x2 + 7x + 3 = (2x + 1)(x + 1) (iii) 6x2 + 5x – 6 We have a = 6, b = 5 and c = –6 ∴ l + m = 5 and lm = ac = 6 × (–6) = –36 ∴ l + m = 9 + (–4) ∴ 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = 3x(2x + 3) – 2(2x + 3) = (2x + 3)(3x – 2) Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2) (iv) 3x2 – x – 4 We have a = 3, b = –1 and c = –4 ∴ l + m = –1 and lm = 3 × (–4) = –12 ∴ l = – 4 and m = 3 Now, 3x2 – x – 4 = 3x2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1) Thus, 3x2 – x – 4 = (3x – 4)(x + 1) 5. Factorise: (i) x3 – 2x2 – x + 2 (ii) x3 – 3x2 – 9x – 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 – 2y – 1 Ans. (i) x3 – 2x2 – x + 2 Rearranging the terms, we have x3 – 2x2 – x + 2 = x3 – x – 2x2 + 2 = x(x2 – 1) – 2(x2 – 1) = (x2 – 1) (x – 2) = [(x2 – (1)2][x – 2] = (x – 1) (x + 1)(x – 2) [∴ a2 – b2 = (a + b)(a – b)] Thus, x3 – 2x2 – x + 2 = (x – 1) (x + 1)(x – 2) (ii) x3 – 3x2 – 9x – 5 We have p(x) = x3 – 3x2 – 9x – 5 By trial, let us find: p(1) = (1)3 – 3(1)2 – 9(1) –5 = 3 – 3 – 9 – 5 = –14 ≠ 0 Now p(–1) = (–1)3 – 3(–1)2 – 9(–1) –5 = –1 – 3(1) + 9 – 5 = –1 – 3 + 9 – 5 = 0 ∴ By factor theorem, [x – (–1)] is a factor of p(x). Now, ∴ x3 – 3x2 – 9x – 5 = (x + 1)(x2 – 4x – 5) = (x + 1)[x2 – 5x + x – 5] [Splitting –4 into –5 and +1] = (x + 1)[x(x – 5) +1(x – 5] = (x + 1)[(x – 5) (x + 1)] = (x + 1)(x – 5)(x + 1) (iii) x3 + 13x2 + 32x + 20 We have p(x) = x3 + 13x2 + 32x + 20 By trial, let us find: p(1) = (1)3 + 13(1)2 + 32(1) + 20 = 1 + 13 + 32 + 20 = 66 ≠ 0 Now p(–1) = (–1)3 + 13(–1)2 + 32(–1) + 20 = –1 + 13 – 32 + 20 = 0 ∴ By factor theorem, [x – (–1)], i.e. (x + 1) is a factor p(x). or x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) = (x + 1) (x2 + 2x + 12x + 20) [Splitting the middle term] = (x + 1)[x(x + 2) + 10(x + 2)] = (x + 1)[(x + 2) (x + 10)] = (x + 1)(x + 2) (x + 10) (iv) 2y3 + y2 – 2y – 1 We have p(y) = 2y3 + y2 – 2y – 1 By trial, we have p(1) = 2(1)3 + (1)2 – 2(1) – 1 = 2(1) + 1 – 2 – 1 = 2 + 1 – 2 – 1 = 0 ∴ By factor theorem, (y – 1) is a factor of p(y). ∴ 2y3 – y2 – 2y – 1 = (y – 1)(2y2 + 3y + 1) = (y – 1)[2y2 + 2y + y + 1) [Splitting the middle term] = (y – 1)[2y(y + 1) + 1(y + 1)] = (y – 1)[(y + 1) (2y + 1)] = (y – 1)(y + 1) (2y + 1) Exercise– 2.5 1. Use suitable identities to find the following products: (i) (x + 4) (x + 10) (ii) (x + 8) (x – 10) (iii) (3x + 4)(3x – 5) (iv) (v) (3 – 2x) (3 + 2x) Ans. (i) (x + 4) (x + 10): Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, we have: (x + 4)(x + 10) = x2 + (14 + 10)x + (4 × 10) = x2 + 14x + 40 (ii) (x + 8) (x – 10): Here, a = 8 and b = (–10) ∴ Using (x + a)(x + b) = x2 + (a + b) + ab, we have: (x + 8)(x – 10) = x2 + [8 + (–10)]x + [8 × (–10)] = x2 + [–2]x + [–80] = x2 + 2x – 80 (iii) (3x + 4)(3x – 5): ∴ Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, we have: (3x + 4)(3x – 5) = (3x)2 + [4 + (–5)]3x + [4 × (–5)] = 9x2 + [–1]3x + [–20] = 9x2 – 3x – 20 (iv) Using the identity (a + b)(a – b) = a2 – b2, we have: (v) (3 – 2x)(3 + 2x): Using the identity (a + b)(a – b) = a2 – b2, we have: (3 – 2x)(3 + 2x) = (3)2 – (2x)2 = 9 – 4x2 2. Evaluate the following products without multiplying directly: (i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96 Ans. (i) We have 103 × 107 = (100 + 3)(100 + 7) = (100)2 + (3 + 7) × 100 + (3 × 7) [Using (x + a)(x + b) = x2 + (a + b)x +ab] = 10000 + (10) × 100 + 21 = 10000 + 1000 + 21 = 11021 (ii) We have 95 × 96 = (100 – 5)(100 – 4) = (100)2 + [(–5) + (–4)] × 100 + [(–5) × (–4)] [Using (x + a)(x + b) = x2 + (a + b)x +ab] = 10000 + [–9] × 100 + 21 = 10000 + (–900) + 20 = 9120 (iii) We have 104 × 96 = (100 + 4)(100 – 4) = (100)2 – (4)2 [Using (a + b)(a – b) = a2 – b2] = 10000 – 16 = 9984 3. Factorise the following using appropriate identities: (i) 9x2 + 6xy + y2 (ii) 4y2 – 4y + 1 (iii) Ans. (i) We have 9x2 + 6xy + y2 = (3x)2 + 2(3x)(y) + (y)2 = (3x + y)2 [Using a2 + 2ab + b2 = (a + b)2] = (3x + y)(3x + y) (ii) We have 4y2 + 4y + 1 = (2y)2 – 2(2y)(1) + (1)2 = (2y – 1)2 [∴ a2 – 2ab + b2 = (a – b)2] = (2y – 1)(2y – 1) [Using a2 – b2 = (a + b)(a – b)] 4. Expand each of the following using suitable identities: (i) (x + 2y + 4z)2 (ii) (2x – y + z)2 (iii) (–2x + 3y + 2z)2 (iv) (3a – 7b – c)2 (v) (–2x + 5y – 3z)2 Ans. (i) (x + 2y + 4z)2 We have (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx ∴ (x + 2y + 4z)2 = (x)2 + (2y)2 + (4y)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x) = x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx (ii) (xx – y + z)2 Using (x + y + z)2= x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (2x – y + z)2 = (2x)2 + (–y)2 + (z)2 + 2(2x)(–y) + 2(–y)(z) + 2(z)(2x) = 4x2 + y2 + z2 – 4xy – 2yz + 4zx (iii) (–2x + 3y + 2z)2 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (–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 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have (3a – 7b – c)2 = (3a)2 + (–7b)2 + (–c)2 + 2(3a)(–7b) + 2(–7b)(–c) + 2(–c)(3a) = 9a2 + 49b2 + c2 + (–42ab) + (14bc) – 6ca = 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca (v) (–2x + 5y – 3z)2 Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have ∴ (–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] = 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx, we have 5. Factorise: Ans. (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz = (2x)2 + (3y)2 + (–4z)2 + 2(2x)(3y) + 2(3y)(–4z) + 2(–4z)(2x) [Using Identity V] = (2x + 3y – 4z)2 = (2x + 3y – 4z)(2x + 3y – 4z) 6. Write the following cubes in expanded form: (i) (2x + 1)3 (ii) 2a – 3b)3 (iii) (iv) Ans. Using Identity VI and Identity VII, we have (x + y)3 = x3 + y3 + 3xy (x + y), and (x + y)3 = x3 + y3 + 3xy (x – y). (i) (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1) [(2x) + (1)] = 8x3 + 1 + 6x[2x + 1] [Using Identity VI] = 8x3 + 1 + 12x2 + 6x = 8x3 + 12x2 + 6x + 1 (ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b) [(2a) + (3b)] = 8a3 – 27b3 – 18ab[2a – 3b] [Using Identity VII] = 8a3 – 27b3 – [36a2b – 54ab2] = 8a3 – 27b3 – 36a2b + 54ab2 7. Evaluate the following using suitable identities: (i) (99)3 (ii) (102)3 (iii) (998)3 Ans. (i) (99)3 We have 99 = 100 – 1 ∴ 993 = (100 – 1)3 = 1003 – 13 – 3(100)(1)(100 – 1) = 1000000 – 1 – 30000 + 300 = 100300 – 30001 = 970299 (ii) (102)3 We have 102 = 100 + 2 (102)3 = (100 + 2)3 = (100)3 + (2)3 + 3(100)(2)[100 2 1] = 1000000 + 8 + 600[100 + 2] = 1000000 + 8 + 60000 + 1200 = 1061208 (ii) (998)3 We have 998 = 100 – 2 ∴ (999)3 = (1000 – 2)3 = (1000)3 – (2)3 – 3(1000)(2)[1000 – 2] = 10000000 – 8 – 6000[1000 – 2] = 10000000 – 8 – 600000 – 12000 = 994011992 8. Factorise each of the following: (i) 8a3 + b3 + 12a2b + 6ab2 (ii) 8a3 – b3 – 12a2b – 6ab2 (iii) 27 – 125a3 – 135a + 225a2 (iv) 64a3 – 27b3 – 144a2b + 108ab2 Ans. (i) 8a3 + b3 + 12a2b + 6ab2 = (2a)3 + (b)3 + 6ab(2a + b) = (2a)3 + (b)3 + 3(2a)(b)(2a + b) = (2a + b)3 [Using Identify VI] = (2a + b)(2a + b)(2a + b) (ii) 8a3 – b3 – 12a2b – 6ab2 = (2a)3 – (b)3 – 3(2a)(b)(2a – b) = (2a – b)3 [Using Identify VII] = (2a – b)(2a – b)(2a – b) (iii) 27 – 125a3 – 135a + 225a2 = (3)3 – (5a)3 – 3(3)(5a)[3 – 5a] = (3 – 5a)3 [Using Identify VII] = (3 – 5a)(3 – 5a)(3 – 5a) (iv) 64a3 – 27b3 – 144a2b + 108ab2 = (4a)3 – (3b)3 – 3(4a)(3b)[4a – 3b] = (4a – 3b)3 [Using Identify VII] = (4a – 3b)(4a – 3b)(4a – 3b) 9. Verify: (i) x3 + y3 = (x + y)(x2 – xy + y2) (ii) x3 – y3 = (x – y)(x2 + xy + y2) Ans. (i) R.H.S. = (x + y)(x2 – xy + y2) = x(x2 – xy + y2) + y(x2 – xy + y2) = x3 – x2y + xy2 + x2y – xy2 + y3 = x3 + y3 = L.H.S (ii) R.H.S. = (x – y)(x2 + xy + y2) = x(x2 + xy + y2) – y(x2 + xy + y2) = x3 + x2y + xy2 – x2y – xy2 – y3 = x3 – y3 = L.H.S 10. Factorise each of the following: (i) 27y3 + 125z3 (ii) 64m3 – 343n3 Remember I. x3 + y3 = (x + y)(x2 + y2 – xy) II. x3 – y3 = (x – y)(x2 + y2 + xy) Ans. (i) Using the identity x3 + y3 = (x + y)(x2 + – xy + y2), we have 27y3 + 125z3 = (3y)3 + (5z)3 = (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2] = (3y + 5z)(9y2 – 15yz + 25z2) (ii) Using the identity x3 – y3 = (x – y)(x2 + xy + y2), we have 64m3 – 343n3 = (4m)3 – (7n)3 = (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2] = (4m + 7n)(16m2 + 28mn + 19n2) 11. Factorise 27x3 + y3 + z3 – 9xyz. Ans. Remember x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) We have 27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z) ∴ Using the identity x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx), we have (3x)3 + y3 +z3 – 3(3x)(y)(z) = (3x + y +z)[(3x)2 + y2 + z2 – (3x × y) – (y × z) – (z × 3x)] = (3x + y +z)(9x2 + y2 + z2 – 3xy – yz – 3zx) 12. Verify that 13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz. Ans. Ans. Since x + y + z = 0 ∴ x + y = –z or (x + y)3 = (–z)3 or x3 + y3 + 3xy(x + y) = –z3 or x3 + y3 + 3xy(–z) = –z3 [ x + y = (–z)] or x3 + y3 – 3xy = –z3 or (x3 + y3 + z3) – 3xy = 0 or (x3 + y3 + z3) = 3xy Hence, If x + y + z = 0, then (x3 + y3 + z3 = 3xy. 14. Without actually calculating the cubes, find the value of each of the following: (i) (–12)3 + (7)3 + (5)3 (ii) (28)3 + (–15)3 + (–13)3 Ans. (i) (–12)3 + (7)3 + (5)3 Let x = –12, y = 7 and z = 5 Then x + y + z = –12 + 7 + 5 = 0 We know that if x + y + z =0, then x3 + y3 + z3 = 3xyz. ∴ (–12)3 + (7)3 + (5)3 = 3[(–12)(7)(5)] [∴ (–12) + 7 + 5 = 0] = 3[–420] = –1260 Thus, (–12)3 + (7)3 + (5)3 = –1260 (ii) (28)3 + (–15)3 + (–13)3 Let x = 28, y = –15 and z = –13 ∴ x + y + z =0, then x3 + y3 + z3 = 3xyz. We know that if x + y + z =0, then x3 + y3 + z3 = 3xyz. (28)3 + (–15)3 + (–13)3 = 3(28)(–15)(–13) [28 + (–15) + (–13) = 0] = 3(5460) = 16380 Thus, (28)3 + (–15)3 + (–13)3 = 1638 15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: Area: 25a – 35a 2 + 12 Area: 35y2 + 13y – 12 (i) (ii) Remember Area of a rectangle = (Length) × (Breadth) Ans. (i) Area = 25a2 – 35a + 12 We have to factorise the polynomial: 25a2 – 35a + 12 Splitting the co-efficient of a, we have –35 = (–20) + (–15) [∴ 25 × 12 = 300 and (–20) × (–15) = 300] = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3) Thus, the possible length and breadth are (5a –3) and (5a – 4). (ii) Area = 35y2 + 13y – 12 We have to factorise the polynomial: 35y2 + 13y – 12 Splitting the middle term, we get 13y = 28y – 15y [∴ 28 × (–15) = –420 and –12 × 35 = –420] ∴ 35y2 + 13y – 12 = 35y2 + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3) Thus, the possible length and breadth are (7y – 3) and (5y + 4). 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume: 3x – 122 x (ii) Volume: 12ky2 + 18ky – 12k Remember Volume of a cuboid = (Length) × (Breadth) × (Height) Ans. (i) Volume = 3x2 – 12x On factorising 3x2 – 12x, we have 3x2 – 12x = 3[x2 – 4x] = 3 × [x(x – 4)] = 3 × x × (x – 4) ∴The possible dimensions of the cuboid are: 3, x and (x – 4) units. (ii) Volume = 12ky2 + 8ky – 20k We have 12ky2 + 8ky – 20k = 4[3ky2 + 2ky – 5k] = 4[2(3y2 + 2y – 5] (Splitting the middle term) = 4k[3y(y – 1) + 5(y – 1)] = 4k[(3y + 5)(y – 1)] = 4k × (3y + 5) × (y – 1) Thus, the possible dimensions are: 4k, (3y + 5) and (y – 1) units. Từ khóa » G(x)=5x^6+x^5+9x^3-12x-125
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