Q1: Use suitable identities to find the following products: (i) (x + 4)(x + 10)
Ans: Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, [Here, a = 4 and b = 10] We get, (x + 4)(x + 10) = x2 + (4 + 10)x + (4 x 10) = x2 + 14x + 40
(ii) (x + 8)(x – 10)
Ans: Using the identity, (x+a)(x+b) = x2+(a+b)x+ab [Here, a = 8 and b = (–10)] We get: (x + 8)(x – 10) = x2 + [8 + (–10)]x + [8 x (–10)] = x2 + [8-10]x + [–80] = x2 – 2x – 80
(iii) (3x + 4)(3x – 5)
Ans: Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, [Here, x = 3x, a = 4 and b = −5] we get (3x + 4)(3x – 5) = (3x)2 + [4 + (–5)]3x + [4 x (–5)] = 9x2 + 3x(4–5)–20 = 9x2 – 3x – 20
(iv) (y2 + 3/2) (y2 – 3/2)
Ans: Using the identity (x + y)(x – y) = x2 – y2, [Here, x = y2 and y = 3/2] we get: (y2+3/2)(y2–3/2) = (y2)2–(3/2)2 = y4–9/4
(v) (3 – 2x) (3 + 2x)
Ans: Using (a + b) (a – b) = a2 – b2, putting a = 3 , b = 2x = (3)2 – (2x)2 = 9 – 4x2
Q2: Evaluate the following products without multiplying directly: (i)103 × 107
Ans: (100+3) × (100+7) Using identity, [(x+a)(x+b) = x2+(a+b)x+ab Here, x = 100 a = 3 b = 7 We get, 103×107 = (100+3)×(100+7) = (100)2+(3+7)100+(3×7)) = 10000+1000+21 = 11021
(ii) 95×96
Ans: (100-5)×(100-4) Using identity, [(x-a)(x-b) = x2-(a+b)x+ab Here, x = 100 a = -5 b = -4 We get, 95×96 = (100-5)×(100-4) = (100)2+100(-5+(-4))+(-5×-4) = 10000-900+20 = 9120
(iii)104 × 96
Ans: (100+4)×(100–4) Using identity, [(a+b)(a-b)= a2-b2] Here, a = 100 b = 4 We get, 104×96 = (100+4)×(100–4) = (100)2–(4)2 = 10000–16 = 9984
Q3: Factorise the following using appropriate identities: (i) 9x2 + 6xy + y2
Ans: (3x)2+(2×3x×y)+y2 Using identity, x2+2xy+y2 = (x+y)2 Here, x = 3x y = y 9x2+6xy+y2 = (3x)2+(2×3x×y)+y2 = (3x+y)2 = (3x+y)(3x+y)
(ii) 4y2 – 4y + 1
Ans: 4y2−4y+1 = (2y)2–(2×2y×1)+1 Using identity, x2 – 2xy + y2 = (x – y)2 Here, x = 2y y = 1 4y2−4y+1 = (2y)2–(2×2y×1)+12 = (2y–1)2 = (2y–1)(2y–1)
(iii) x2– y2/100
Ans: x2–y2/100 = x2–(y/10)2 Using identity, x2-y2 = (x-y)(x+y) Here, x = x y = y/10 x2–y2/100 = x2–(y/10)2 = (x–y/10)(x+y/10)
Q4: Expand each of the following, using suitable identities: (i) (x + 2y + 4z)2
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = x y = 2y z = 4z (x+2y+4z)2 = x22+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x) = x2+4y2+16z2+4xy+16yz+8xz
(ii) (2x – y + z)2
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = 2x y = −y z = z (2x−y+z)2 = (2x)2+(−y)2+z2+(2×2x×−y)+(2×−y×z)+(2×z×2x) = 4x2+y2+z2–4xy–2yz+4xz
(iii) (–2x + 3y + 2z)2
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = −2x y = 3y z = 2z (−2x+3y+2z)2 = (−2x)2+(3y)2+(2z)2+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x) = 4x2+9y2+4z2–12xy+12yz–8xz
(iv) (3a – 7b – c)2
Ans: Using identity (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = 3a y = – 7b z = – c (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
(v) (–2x + 5y – 3z)2
Ans: Using identity, (x+y+z)2= x2+y2+z2+2xy+2yz+2zx Here, x = –2x y = 5y z = – 3z (–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
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = (1/4)a y = (-1/2)b z = 1
Q5: Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2 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 = (2x+3y–4z)(2x+3y–4z)
(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2 yz – 8xz
Ans: Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2 2x2+y2+8z2–2√2xy+4√2yz–8xz = (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x) = (−√2x+y+2√2z)2 = (−√2x+y+2√2z)(−√2x+y+2√2z)
Q6: Write the following cubes in expanded form: (i) (2x + 1)3
Ans: Using identity,(x+y)3 = x3+y3+3xy(x+y) ((3/2)x+1)3=((3/2)x)3+13+(3×(3/2)x×1)((3/2)x +1)
(iv) (x−(2/3)y)3
Ans: Using identity, (x –y)3 = x3–y3–3xy(x–y)
Q7: Evaluate the following using suitable identities: (i) (99)3
Ans: We can write 99 as 100–1 Using identity, (x –y)3 = x3–y3–3xy(x–y) (99)3 = (100–1)3 = (100)3–13–(3×100×1)(100–1) = 1000000 –1–300(100 – 1) = 1000000–1–30000+300 = 970299 = 970299
(ii) (102)3
Ans: We can write 102 as 100+2 Using identity,(x+y)3 = x3+y3+3xy(x+y) (100+2)3 =(100)3+23+(3×100×2)(100+2) = 1000000 + 8 + 600[100 + 2] = 1000000 + 8 + 60000 + 1200 = 1061208
(iii) (998)3
Ans: We can write 99 as 1000–2 Using identity,(x–y)3 = x3–y3–3xy(x–y) (998)3 =(1000–2)3 =(1000)3–23–(3×1000×2)(1000–2) = 1000000000–8–6000(1000– 2) = 1000000000–8- 6000000+12000 = 994011992
Q8: Factorise each of the following: (i) 8a3 + b3 + 12a2b + 6ab2
Ans: The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2 8a3+b3+12a2b+6ab2 = (2a)3+b3+3(2a)2b+3(2a)(b)2 = (2a+b)3 = (2a+b)(2a+b)(2a+b) Here, the identity, (x +y)3 = x3+y3+3xy(x+y) is used.
(ii) 8a3 – b3 – 12a2b + 6ab2
Ans: The expression, 8a3–b3−12a2b+6ab2 can be written as (2a)3–b3–3(2a)2b+3(2a)(b)2 8a3–b3−12a2b+6ab2 = (2a)3–b3–3(2a)2b+3(2a)(b)2 = (2a–b)3 = (2a–b)(2a–b)(2a–b) Here, the identity,(x–y)3 = x3–y3–3xy(x–y) is used.
(iii) 27 – 125a3 – 135a + 225a2
Ans: The expression, 27–125a3–135a +225a2 can be written as 33–(5a)3–3(3)2(5a)+3(3)(5a)2 27–125a3–135a+225a2 = 33–(5a)3–3(3)2(5a)+3(3)(5a)2 = (3–5a)3 = (3–5a)(3–5a)(3–5a) Here, the identity, (x–y)3 = x3–y3-3xy(x–y) is used.
(iv) 64a3 – 27b3 – 144 a2b + 108 ab2
Ans: The expression, 64a3–27b3–144a2b+108ab2 can be written as (4a)3–(3b)3 – 3(4a)2(3b)+3(4a)(3b)2 64a3–27b3 – 144a2b+108ab2 = (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2 =(4a–3b)3 =(4a–3b)(4a–3b)(4a–3b) Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.
(v) 27p3– (1/216)−(9/2) p2+(1/4)p
Ans: The expression, 27p3–(1/216)−(9/2) p2+(1/4)p can be written as (3p)3–(1/6)3−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6) Using (x – y)3 = x3 – y3 – 3xy (x – y) 27p3–(1/216)−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6) Taking x = 3p and y = 1/6 = (3p–1/6)3 = (3p–1/6)(3p–1/6)(3p–1/6)
Q10: Factorise each of the following: (i) 27y3+125z3
Ans: The expression, 27y3+125z3 can be written as (3y)3+(5z)3 27y3+125z3 = (3y)3+(5z)3 We know that, x3+y3 = (x+y)(x2–xy+y2) 27y3+125z3 = (3y)3+(5z)3 = (3y+5z)[(3y)2–(3y)(5z)+(5z)2] = (3y+5z)(9y2–15yz+25z2)
(ii) 64m3–343n3
Ans: The expression, 64m3–343n3can be written as (4m)3–(7n)3 64m3–343n3 = (4m)3–(7n)3 We know that, x3–y3 = (x–y)(x2+xy+y2) 64m3–343n3 = (4m)3–(7n)3 = (4m-7n)[(4m)2+(4m)(7n)+(7n)2] = (4m-7n)(16m2+28mn+49n2)
Q11: Factorise 27x3 + y3 + z3 – 9xyz.
Solution: The expression27x3+y3+z3–9xyz can be written as (3x)3+y3+z3–3(3x)(y)(z) 27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z) We know that, 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+z2–3xy–yz–3xz] = (3x+y+z)(9x2+y2+z2–3xy–yz–3xz)
Q12: Verify that x3 + y3 + z3 – 3xyz = (1/2) (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]
Q13: If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Ans: We know that, x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2–xy–yz–xz) Now, according to the question, let (x+y+z) = 0, then, x3+y3+z3 -3xyz = (0)(x2+y2+z2–xy–yz–xz) ⇒ x3+y3+z3–3xyz = 0 ⇒x3+y3+z3 = 3xyz Hence Proved
Q14: Without actually calculating the cubes, find the value of each of the following: (i) (–12)3 + (7)3 + (5)3
Ans: Let a = −12 b = 7 c = 5 We know that if x+y+z = 0, then x3+y3+z3=3xyz. Here, −12+7+5=0 (−12)3+(7)3+(5)3 = 3xyz = 3×-12×7×5 = -1260
(ii) (28)3 + (–15)3 + (–13)3
Ans: Let a = 28 b = −15 c = −13 We know that if x+y+z = 0, then x3+y3+z3 = 3xyz. Here, x+y+z = 28–15–13 = 0 (28)3+(−15)3+(−13)3 = 3xyz = 0+3(28)(−15)(−13) = 16380
Q15: Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : 25a2–35a+12
Ans: Using the splitting the middle term method, We have to find a number whose sum = -35 and product =25×12=300 We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20=300] 25a2–35a+12 = 25a2–15a−20a+12 = 5a(5a–3)–4(5a–3) = (5a–4)(5a–3) Possible expression for length = 5a–4 Possible expression for breadth = 5a –3
(ii) Area : 35y2+13y–12
Ans: Using the splitting the middle term method, We have to find a number whose sum = 13 and product = 35×-12 = 420 We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420] 35y2+13y–12 = 35y2–15y+28y–12 = 5y(7y–3)+4(7y–3) = (5y+4)(7y–3) Possible expression for length = (5y+4) Possible expression for breadth = (7y–3)
Q16: What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume: 3x2–12x
Ans: 3x2–12x can be written as 3x(x–4) by taking 3x out of both the terms. Possible expression for length = 3 Possible expression for breadth = x Possible expression for height = (x–4)
(ii) Volume: 12ky2+8ky–20k
Ans: 12ky2+8ky–20k can be written as 4k(3y2+2y–5) by taking 4k out of both the terms. 12ky2+8ky–20k = 4k(3y2+2y–5) [Here, 3y2+2y–5 can be written as 3y2+5y–3y–5 using splitting the middle term method.] = 4k(3y2+5y–3y–5) = 4k[y(3y+5)–1(3y+5)] = 4k(3y+5)(y–1) Possible expression for length = 4k Possible expression for breadth = (3y +5) Possible expression for height = (y -1)
Q1. Determine which of the following polynomials has (x + 1) a factor: (i) x3 + x2 + x + 1 Ans: Let p(x) = x3 + x2 + x + 1 The zero of x + 1 is -1. [x + 1 = 0 means x = -1] p(−1) = (−1)3 + (−1)2 + (−1) + 1 = −1 + 1 − 1 + 1 = 0 ∴ By factor theorem, x + 1 is a factor of x3 + x2 + x + 1
(ii) x4 + x3 + x2 + x + 1 Ans: Let p(x) = x4 + x3 + x2 + x + 1 The zero of x + 1 is -1. . [x + 1= 0 means x = -1] p(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1 = 1 − 1 + 1 − 1 + 1 = 1 ≠ 0 ∴ By factor theorem, x+1 is not a factor of x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1 Ans: Let p(x)= x4 + x3 + x2 + x + 1 The zero of x+1 is -1. p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1 =1−3+3−1+1 =1 ≠ 0 ∴ By factor theorem, x + 1 is not a factor of x4 +3x3 + 3x2 + x + 1
(iv) x3 – x2– (2+√2)x +√2 Ans: Let p(x) = x3–x2–(2+√2)x +√2 The zero of x+1 is -1. p(−1) = (-1)3–(-1)2–(2+√2)(-1) + √2 = −1−1+2+√2+√2 = 2√2 ≠ 0 ∴ By factor theorem, x+1 is not a factor of x3–x2–(2+√2)x +√2
Q2. 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 Ans: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴ Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2 + 1 + 2 − 1 = 0 ∴ By factor theorem, g(x) is a factor of p(x).
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 Ans: p(x) = x3+ 3x2 3x + 1, g(x) = x + 2 g(x) = 0 ⇒ x + 2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8 + 12 − 6 + 1 = −1 ≠ 0 ∴ By factor theorem, g(x) is not a factor of p(x).
(iii) p(x)= x3 – 4x2 + x + 6, g(x) = x – 3 Ans: p(x) = x3– 4x2 + x + 6, g(x) = x – 3 g(x) = 0 ⇒ x−3 = 0 ⇒ x = 3 ∴ Zero of g(x) is 3. Now, p(3) = (3)3−4(3)2 + (3) + 6 = 27 − 36 + 3 + 6 = 0 ∴ By factor theorem, g(x) is a factor of p(x).
Q3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases: (i) p(x) = x2 + x + k Ans: If x – 1 is a factor of p(x), then p(1) = 0 By Factor Theorem ⇒ (1)2+(1)+k = 0 ⇒ 1+1+k = 0 ⇒ 2+k = 0 ⇒ k = −2
(ii) p(x) = 2x2 + kx + √2 Ans: If x-1 is a factor of p(x), then p(1)=0 ⇒ 2(1)2 + k(1) + √2 = 0 ⇒ 2 + k + √2 = 0 ⇒ k = −(2 + √2)
(iii) p(x) = kx2–√2x + 1 Ans: If x – 1 is a factor of p(x), then p(1)=0 By Factor Theorem ⇒ k(1)2-√2(1)+1=0 ⇒ k = √2-1
(iv) p(x) = kx2 – 3x + k Ans: If x-1 is a factor of p(x), then p(1) = 0 By Factor Theorem ⇒ k(1)2–3(1)+k = 0 ⇒ k−3+k = 0 ⇒ 2k−3 = 0 ⇒ k= 3/2
Q4. Factorize: (i) 12x2 – 7x + 1 Ans: Using the splitting the middle term method, We have to find a number whose sum = -7 and product =1×12 = 12 We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12] 12x2–7x+1= 12x2-4x-3x+1 = 4x(3x-1)-1(3x-1) = (4x-1)(3x-1)
(ii) 2x2 + 7x + 3 Ans: Using the splitting the middle term method, We have to find a number whose sum = 7 and product = 2×3 = 6 We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6] 2x2+7x+3 = 2x2+6x+1x+3 = 2x (x+3)+1(x+3) = (2x+1)(x+3)
(iii) 6x2 + 5x – 6 Ans: Using the splitting the middle term method, We have to find a number whose sum = 5 and product = 6×-6 = -36 We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36] 6x2+5x-6 = 6x2+9x–4x–6 = 3x(2x+3)–2(2x+3) = (2x+3)(3x–2)
(iv) 3x2–x–4 Ans: Using the splitting the middle term method, We have to find a number whose sum = -1 and product = 3 × -4 = -12 We get -4 and 3 as the numbers [-4 + 3 = -1 and -4 × 3 = -12] 3x2 – x – 4 = 3x2 – x – 4 = 3x2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
5. Factorize: (i) x3– 2x2 – x + 2 Ans: Let p(x) = x3–2x2–x+2 Factors of 2 are ±1 and ± 2 Now, p(x) = x3–2x2–x+2 p(−1) = (−1)3–2(−1)2–(−1)+2 = −1−2+1+2 = 0 Therefore, (x+1) is the factor of p(x)
Now, Dividend = Divisor × Quotient + Remainder Now by splitting the middle term method, (x+1)(x2–3x+2) = (x+1)(x2–x–2x+2) = (x+1)(x(x−1)−2(x−1)) = (x+1)(x−1)(x-2)
(ii) x3 – 3x2 – 9x – 5 Ans: Let p(x) = x3–3x2–9x–5 Factors of 5 are ±1 and ±5 By trial method, we find that p(5) = 0 So, (x-5) is factor of p(x) Now, p(x) = x3–3x2–9x–5 p(5) = (5)3–3(5)2–9(5)–5 = 125−75−45−5 = 0 Therefore, (x-5) is the factor of p(x)
(iii) x3 + 13x2 + 32x + 20 Ans: Let p(x) = x3+13x2+32x+20 Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20 By trial method, we find that p(-1) = 0 So, (x+1) is factor of p(x) Now, p(x)= x3+13x2+32x+20 p(-1) = (−1)3+13(−1)2+32(−1)+20 = −1+13−32+20 = 0 Therefore, (x+1) is the factor of p(x)
Now, Dividend = Divisor × Quotient +Remainder Now by splitting the middle term method, (x+1)(x2+12x+20) = (x+1)(x2+2x+10x+20) = (x−5)x(x+2)+10(x+2) = (x−5)(x+2)(x+10)
(iv) 2y3 + y2 – 2y – 1 Ans: Let p(y) = 2y3+y2–2y–1 Factors = 2×(−1)= -2 are ±1 and ±2 By trial method, we find that p(1) = 0 So, (y-1) is factor of p(y) Now, p(y) = 2y3+y2–2y–1 p(1) = 2(1)3+(1)2–2(1)–1 = 2+1−2 = 0 Therefore, (y-1) is the factor of p(y)
Now, Dividend = Divisor × Quotient + Remainder Now by splitting the middle term method, (y−1)(2y2+3y+1) = (y−1)(2y2+2y+y+1) = (y−1)(2y(y+1)+1(y+1)) = (y−1)(2y+1)(y+1)
Q3. Verify whether the following are zeroes of the polynomial, indicated against them. (i) p(x) = 3x + 1, x = −1/3 Ans: For, x = -1/3, p(x) = 3x+1 ∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0 ∴ -1/3 is a zero of p(x).
(ii) p(x) = 5x – π, x = 4/5 Ans: For, x = 4/5, p(x) = 5x – π ∴ p(4/5) = 5(4/5) – π = 4 -π ∴ 4/5 is not a zero of p(x).
(iii) p(x) = x2 − 1, x = 1, −1 Ans: For, x = 1, −1; p(x) = x2−1 ∴ p(1) = 12 − 1 = 1 − 1 = 0 p(−1) = (-1)2 − 1 = 1 − 1 = 0 ∴ 1, −1 are zeros of p(x).
(iv) p(x) = (x+1)(x–2), x =−1, 2 Ans: For, x = −1,2; p(x) = (x+1)(x–2) ∴ p(−1) = (−1+1)(−1–2) = (0)(−3) = 0 p(2) = (2+1)(2–2) = (3)(0) = 0 ∴ −1,2 are zeros of p(x).
(v) p(x) = x2, x = 0 Ans: For, x = 0 p(x) = x2 p(0) = 02 = 0 ∴ 0 is a zero of p(x).
(vi) p(x) = lx + m, x = −m/l Ans: For, x = -m/l ; p(x) = lx+m ∴ p(-m/l)= l(-m/l)+m = −m+m = 0 ∴ -m/l is a zero of p(x).
(vii) p(x) = 3x2−1, x = -1/√3 , 2/√3 Ans: For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1 ∴ p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 0 ∴ p(2/√3 ) = 3(2/√3)2-1 = 3(4/3)-1 = 4−1=3 ≠ 0 ∴ -1/√3 is a zero of p(x) but 2/√3 is not a zero of p(x).
(viii) p(x) =2x + 1, x = 1/2 Ans: For, x = 1/2 p(x) = 2x + 1 ∴ p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0 ∴ 1/2 is not a zero of p(x).
Q4. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 Ans: p(x) = x + 5 ⇒ x + 5 = 0 ⇒ x = −5 ∴ -5 is a zero polynomial of the polynomial p(x).
(ii) p(x) = x – 5 Ans: p(x) = x − 5 ⇒ x − 5 = 0 ⇒ x = 5 ∴ 5 is a zero polynomial of the polynomial p(x).
(iii) p(x) = 2x + 5 Ans: p(x) = 2x + 5 ⇒ 2x+5 = 0 ⇒ 2x = −5 ⇒ x = -5/2 ∴ x = -5/2 is a zero polynomial of the polynomial p(x).
(iv) p(x) = 3x–2 Ans: p(x) = 3x–2 ⇒ 3x − 2 = 0 ⇒ 3x = 2 ⇒x = 2/3 ∴ x = 2/3 is a zero polynomial of the polynomial p(x).
(v) p(x) = 3x Ans: p(x) = 3x ⇒ 3x = 0 ⇒ x = 0 ∴ 0 is a zero polynomial of the polynomial p(x).
(vi) p(x) = ax, a ≠ 0 Ans: p(x) = ax ⇒ ax = 0 ⇒ x = 0 ∴ x = 0 is a zero polynomial of the polynomial p(x).
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. Ans: p(x) = cx + d ⇒ cx + d =0 ⇒ x = -d/c ∴ x = -d/c is a zero polynomial of the polynomial p(x).
Q1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x2–3x+7 Ans: The equation 4x2–3x+7 can be written as 4x2 – 3x1 + 7x0 Since x is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers, we can say that the expression 4x2 – 3x + 7 is a polynomial in one variable.
(ii) y2+√2 Ans: The equation y2 + √2 can be written as y2 + √2y0 Since y is the only variable in the given equation and the powers of y (i.e., 2 and 0) are whole numbers, we can say that the expression y2 + √2 is a polynomial in one variable.
(iii) 3√t + t√2 Ans: The equation 3√t + t√2 can be written as 3t1/2 + √2t Though t is the only variable in the given equation, the powers of t (i.e.,1/2) is not a whole number. Hence, we can say that the expression 3√t + t√2 is not a polynomial in one variable.
(iv) y + 2/y Ans: The equation y + 2/y can be written as y + 2y-1 Though y is the only variable in the given equation, the powers of y (i.e.,-1) is not a whole number. Hence, we can say that the expression y + 2/y is not a polynomial in one variable.
(v) x10 + y3 + t50 Ans: Here, in the equation x10 + y3 + t50 Though the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression x10 + y3 + t50. Hence, it is not a polynomial in one variable.
Q2. Write the coefficients of x2 in each of the following: (i) 2 + x2 + x Ans: The equation 2 + x2+x can be written as 2 + (1)x2 + x We know that, coefficient is the number which multiplies the variable. Here, the number that multiplies the variable x2 is 1 the coefficients of x2 in 2 + x2 + x is 1.
(ii) 2 – x2 + x3 Ans: The equation 2 – x2 + x3 can be written as 2 + (–1)x2 + x3 We know that, coefficient is the number (along with its sign, i.e., – or +) which multiplies the variable. Here, the number that multiplies the variable x2 is -1 the coefficients of x2 in 2 – x2 + x3 is -1.
(iii) (π/2)x2 + x Ans: The equation (π/2)x2 + x can be written as (π/2)x2 + x We know that, coefficient is the number (along with its sign, i.e., – or +) which multiplies the variable. Here, the number that multiplies the variable x2 is π/2. the coefficients of x2 in (π/2)x2 +x is π/2.
(iv)√2x – 1 Ans: The equation √2x – 1 can be written as 0x2+√2x-1 [Since 0x2 is 0] We know that, coefficient is the number (along with its sign, i.e., – or +) which multiplies the variable. Here, the number that multiplies the variable x2 is 0, the coefficients of x2 in √2x – 1 is 0.
Q3. Give one example each of a binomial of degree 35, and of a monomial of degree 100. Ans: The degree of a polynomial is the highest power of the variable in the polynomial. It represents the highest exponent of the variable within the algebraic expression. Therefore,
Binomial of degree 35: A polynomial having two terms and the highest degree 35 is called a binomial of degree 35 Example: 3x35+5
Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of degree 100 Example: 4x100
Q4. Write the degree of each of the following polynomials: (i) 5x3 + 4x2 + 7x Ans: The highest power of the variable in a polynomial is the degree of the polynomial. Here, 5x3 + 4x2 + 7x = 5x3 + 4x2 + 7x1 The powers of the variable x are: 3, 2, 1 The degree of 5x3 + 4x2 + 7x is 3 as 3 is the highest power of x in the equation.
(ii) 4 – y2 Ans: The highest power of the variable in a polynomial is the degree of the polynomial. Here, in 4–y2, The power of the variable y is 2 The degree of 4 – y2 is 2 as 2 is the highest power of y in the equation.
(iii) 5t – √7 Ans: The highest power of the variable in a polynomial is the degree of the polynomial. Here, in 5t–√7, The power of the variable t is 1. The degree of 5t–√7 is 1 as 1 is the highest power of y in the equation.
(iv) 3 Ans: The highest power of the variable in a polynomial is the degree of the polynomial. Here, 3 = 3 × 1 = 3 × x0 The power of the variable here is: 0 The degree of 3 is 0.
Q5. Classify the following as linear, quadratic and cubic polynomials: Ans: We know that, Linear polynomial: A polynomial of degree one is called a linear polynomial. Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial. Cubic polynomial: A polynomial of degree three is called a cubic polynomial. (i) x2 + x Ans: The highest power of x2 + x is 2 The degree is 2 Hence, x2 + x is a quadratic polynomial
(ii) x – x3 Ans: The highest power of x–x3 is 3 The degree is 3 Hence, x–x3 is a cubic polynomial
(iii) y + y2 + 4 Ans: The highest power of y+y2+4 is 2 The degree is 2 Hence, y+y2+4is a quadratic polynomial
(iv) 1 + x Ans: The highest power of 1 + x is 1 The degree is 1 Hence, 1 + x is a linear polynomial.
(v) 3t Ans: The highest power of 3t is 1 The degree is 1 Hence, 3t is a linear polynomial.
(vi) r2 Ans: The highest power of r2 is 2 The degree is 2 Hence, r2 is a quadratic polynomial.
(vii) 7x3 Ans: The highest power of 7x3 is 3 The degree is 3 Hence, 7x3 is a cubic polynomial.
Q1. Write the following in decimal form and mention the kind of decimal expansion each has. (i) 36/100 Ans:
= 0.36 (Terminating)
(ii) 1/11 Ans:
(iii) Ans:
= 4.125 (Terminating)
(iv) 3/13 Ans:
(v) 2/11 Ans:
(vi) 329/400 Ans:
= 0.8225 (Terminating)
Q2. You know that 1/7 = . Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, and 6/7 are, without actually doing the long division? If so, how? [Hint: Study the remainders while finding the value of 1/7 carefully.] Ans:
Q3. Express the following in the form p/q, where p and q are integers and q ≠ 0. (i) Ans:
Assume that x = 0.666… Then,10x = 6.666… 10x = 6 + x 9x = 6 x = 2/3
(ii) Ans:
= (4/10) + (0.777/10) Assume that x = 0.777… Then, 10x = 7.777… 10x = 7 + x x = 7/9 (4/10) + (0.777../10) = (4/10) + (7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9 × 10) = 7/90) = (36/90) + (7/90) = 43/90
(iii) Ans:
Assume that x = 0.001001… Then, 1000x = 1.001001… 1000x = 1 + x 999x = 1 x = 1/999
Q4. Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. Ans: Assume that x = 0.9999…..Eq (a) Multiplying both sides by 10, 10x = 9.9999…. Eq. (b) Eq.(b) – Eq.(a), we get (10x = 9.9999)-(x = 0.9999…) 9x = 9 x = 1 The difference between 1 and 0.999999 is 0.000001 which is negligible. Hence, we can conclude that, 0.999 is very close to 1, therefore, 1 as the answer can be justified.
Q5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer. Ans: 1/17 Dividing 1 by 17:
There are 16 digits in the repeating block of the decimal expansion of 1/17.
Q6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy? Ans: We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is a terminating one. For example: 1/2 = 0. 5, denominator q = 21 7/8 = 0. 875, denominator q =23 4/5 = 0. 8, denominator q = 51 We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.
Q7. Write three numbers whose decimal expansions are non-terminating and non-recurring. Ans: We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating and non-recurring are:
√3 = 1.732050807568
√26 =5.099019513592
√101 = 10.04987562112
Q8. Find three different irrational numbers between the rational numbers 5/7 and 9/11. Ans:
Three different irrational numbers are:
0.73073007300073000073…
0.75075007300075000075…
0.76076007600076000076…
Q9. Classify the following numbers as rational or irrational according to their type: (i)√23 Ans: √23 = 4.79583152331… Since the number is non-terminating and non-recurring, therefore, it is an irrational number.
(ii)√225 Ans: √225 = 15 = 15/1 Since the number can be represented in p/q form, it is a rational number.
(iii) 0.3796 Ans: Since the number, 0.3796, is terminating, it is a rational number.
(iv) 7.478478 Ans: The number, 7.478478, is non-terminating but recurring, it is a rational number.
(v) 1.101001000100001… Ans: Since the number, 1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.
Exercise 1.4
Q1. Classify the following numbers as rational or irrational: (i) 2 –√5 Ans: We know that, √5 = 2.2360679… Here, 2.2360679…is non-terminating and non-recurring. Now, substituting the value of √5 in 2 –√5, we get, 2-√5 = 2-2.2360679… = -0.2360679 Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.
(ii) (3 +√23)- √23 Ans: (3 +√23) –√23 = 3+√23–√23 = 3 = 3/1 Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational.
(iii) 2√7/7√7 Ans: 2√7/7√7 = ( 2/7) × (√7/√7) We know that (√7/√7) = 1 Hence, ( 2/7) × (√7/√7) = (2/7) × 1 = 2/7 Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.
(iv) 1/√2 Ans: Multiplying and dividing the numerator and denominator by √2 we get, (1/√2) × (√2/√2)= √2/2 ( since √2 × √2 = 2) We know that, √2 = 1.4142… Then, √2/2 = 1.4142/2 = 0.7071.. Since the number 0.7071..is non-terminating and non-recurring, 1/√2 is an irrational number.
(v) 2 Ans: We know that, the value of = 3.1415 Hence, 2 = 2 × 3.1415.. = 6.2830… Since the number, 6.2830…, is non-terminating and non-recurring, 2 is an irrational number.
Q2. Simplify each of the following expressions: (i) (3 + √3) (2 + √2) Ans: (3 + √3) (2 + √2 ) Opening the brackets, we get, (3 × 2) + (3 × √2) + (√3 × 2) + (√3 × √2) = 6 + 3√2 + 2√3 + √6
Q3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction? Ans: There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…
Q4. Represent (√9.3) on the number line. Ans: Step 1: Draw a 9.3-unit long line segment, AB. Extend AB to C such that BC = 1 unit. Step 2: Now, AC = 10.3 units. Let the centre of AC be O. Step 3: Draw a semi-circle of radius OC with centre O. Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD. Step 5: OBD, obtained, is a right-angled triangle. Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1 OB = OC – BC ⟹ (10.3/2)-1 = 8.3/2 Using Pythagoras theorem, We get, OD2 = BD2 + OB2 ⟹ (10.3/2)2 = BD2+(8.3/2)2 ⟹ BD2 = (10.3/2)2-(8.3/2)2 ⟹ (BD)2 = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2) ⟹ BD2 = 9.3 ⟹ BD = √9.3 Thus, the length of BD is √9.3. Step 6: Taking BD as the radius and B as the centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.
Q5. Rationalize the denominators of the following: (i) 1/√7 Ans: Multiply and divide 1/√7 by √7 (1×√7)/(√7×√7) = √7/7
(ii) 1/(√7-√6) Ans: Multiply and divide 1/(√7 – √6) by (√7 + √6) [1/(√7 – √6)] × (√7 + √6)/(√7 + √6) = (√7 + √6)/(√7 – √6)(√7 + √6) = (√7 + √6)/√72 – √62 [denominator is obtained by the property, (a + b)(a – b) = a2 – b2] = (√7 + √6)/(7 – 6) = (√7 + √6)/1 = √7 + √6
(iii) 1/(√5+√2) Ans: Multiply and divide 1/(√5 + √2) by (√5 – √2) [1/(√5 + √2)] × (√5 – √2)/(√5 – √2) = (√5 – √2)/(√5 + √2)(√5 – √2) = (√5 – √2)/(√52 – √22) [denominator is obtained by the property, (a + b)(a – b) = a2 – b2] = (√5 – √2)/(5 – 2) = (√5 – √2)/3
(iv) 1/(√7-2) Ans: Multiply and divide 1/(√7 – 2) by (√7 + 2) 1/(√7 – 2) × (√7 + 2)/(√7 + 2) = (√7 + 2)/(√7 – 2)(√7 + 2) = (√7 + 2)/(√72 – 22) [denominator is obtained by the property, (a + b)(a – b) = a2-b2] = (√7 + 2)/(7 – 4) = (√7 + 2)/3
Q1. State whether the following statements are true or false. Justify your answers. (i) Every irrational number is a real number. Ans: True
Reason: Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0. i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000…. Real numbers – The collection of both rational and irrational numbers are known as real numbers. i.e., Real numbers = √2, √5, 56 , 0.102… Every irrational number is a real number, however, every real number is not an irrational number.
(ii) Every point on the number line is of the form √m where m is a natural number. Ans: False
Reason: The statement is false because as per the rule, a negative number cannot be expressed as square roots. E.g., √9 =3 is a natural number. But √2 = 1.414 is not a natural number. Similarly, we know that negative numbers exist on the number line, but their square roots are not real numbers; they are complex. E.g., √-7 = 7i, where i = √-1 The statement that every point on the number line is of the form √m, where m is a natural number is false.
(iii) Every real number is an irrational number. Ans: False
Reason: The statement is false, the real numbers include both irrational and rational numbers. Therefore, every real number cannot be an irrational number. Real numbers – The collection of both rational and irrational numbers are known as real numbers. i.e., Real numbers = √2, √5, , 0.102… Irrational Numbers – A number is said to be irrational if it cannot be written in the p/q, where p and q are integers and q ≠ 0. i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000…. Every irrational number is a real number, however, every real number is not irrational.
Q2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number. Ans: No, the square roots of all positive integers are not irrational. For example, √4 = 2 is rational. √9 = 3 is rational. Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).
Q3. Show how √5 can be represented on the number line. Ans: Step 1: Let line AB be of 2 unit on a number line. Step 2: At B, draw a perpendicular line BC of length 1 unit. Step 3: Join CA Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem, AB2 + BC2 = CA2 22 + 12 = CA2 = 5 ⇒ CA = √5 . Thus, CA is a line of length √5 unit. Step 5: Taking CA as a radius and A as a center draw an arc touching the number line. The point at which number line get intersected by arc is at √5 distance from 0 because it is a radius of the circle whose center was A. Thus, √5 is represented on the number line as shown in the figure.
√5 on number line Q4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1 P2 perpendicular to OP1 of unit length (see Fig. 1.9). Now draw a line segment P2 P3 perpendicular to OP2 . Then draw a line segment P3 P4 perpendicular to OP3. Continuing in this manner, you can get the line segment Pn–1Pn by drawing a line segment of unit length perpendicular to OPn–1. In this manner, you will have created the points P2 , P3 ,…., Pn ,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …. Ans: Step 1: Mark a point O on the paper. Here, O will be the center of the square root spiral. Step 2: From O, draw a straight line, OA, of 1cm horizontally. Step 3: From A, draw a perpendicular line, AB, of 1 cm. Step 4: Join OB. Here, OB will be of √2 Step 5: Now, from B, draw a perpendicular line of 1 cm and mark the end point C. Step 6: Join OC. Here, OC will be of √3 Step 7: Repeat the steps to draw √4, √5, √6….
Q1. Is zero a rational number? Can you write it in the form p/q where p and q are integers and q ≠ 0? Ans: We know that a number is said to be rational if it can be written in the form p/q , where p and q are integers and q
Taking the case of ‘0’,
Yes, zero is a rational number. 0 = Example :
Since, it satisfies the necessary condition, we can conclude that 0 can be written in the p/q form, where q can either be positive or negative number. Hence, 0 is a rational number.
Q2. Find six rational numbers between 3 and 4. Ans: We can find any number of rational numbers between two rational numbers. There are infinite rational numbers between 3 and 4.
First of all, we make the denominators same by multiplying or dividing the given rational numbers by a suitable number.
If denominator is already same then depending on number of rational numbers we need to find in question, we add one and multiply the result by numerator and denominator.
As we have to find 6 rational numbers between 3 and 4, we will multiply both the numbers, 3 and 4, with 6+1 = 7 (or any number greater than 6)
We can choose 6 rational numbers as: and
Q3. Find five rational numbers between 3/4 and 4/5. Ans: There are infinite rational numbers between 3/4 and 4/5.
Since we need to make the denominators same first, then
To find out 5 rational numbers between 3/5 and 4/5, we will multiply both the numbers 3/5 and 4/5
with 5+1=6 (or any number greater than 5)
∴ Five rational numbers between areand
Q4. State whether the following statements are true or false. Give reasons for your answers. (i) Every natural number is a whole number. Ans: True Natural Numbers: Natural numbers are set of numbers that contain numbers from 1 to infinity. Set of natural numbers is represented as N= {1, 2, 3…….}. Whole numbers: Numbers starting from 0 to infinity (without fractions or decimals) Set of Whole numbers is represented as W= 0,1,2,3…. As all the natural numbers comes in the set of whole numbers. Hence, every natural number is a whole number.
(ii) Every integer is a whole number. Ans: False Integers: Integers are set of numbers that contain positive, negative and 0; excluding fractional and decimal numbers. i.e., integers = {…-4,-3,-2,-1,0,1,2,3,4…} Whole numbers: Numbers starting from 0 to infinity (without fractions or decimals) i.e., Whole numbers = 0,1,2,3…. Hence, we can say that integers include whole numbers as well as negative numbers. Every whole number is an integer; however, every integer is not a whole number.
(iii) Every rational number is a whole number. Ans: False Rational numbers: All numbers in the form p/q, where p and q are integers and q ≠ 0. i.e., Rational numbers = 0, 19/30 , 2, 9/-3, -12/7… Whole numbers: Numbers starting from 0 to infinity (without fractions or decimals) i.e., Whole numbers = 0,1,2,3…. Hence, we can say that integers include whole numbers as well as negative numbers. Every whole number is rational, however, every rational number is not a whole number.
