Rs Aggarwal 2019 Solutions for Class 9 Math Chapter 3 Factorisation Of Polynomials are provided here with simple step-by-step explanations. These solutions for Factorisation Of Polynomials are extremely popular among Class 9 students for Math Factorisation Of Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 9 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.
Page No 99:
Question 1:
Factorize:
9x2 + 12xy
Answer:
We have:
Page No 99:
Question 2:
Factorize:
18x2y − 24xyz
Answer:
We have:
Page No 99:
Question 3:
Factorize:
27a3b3 − 45a4b2
Answer:
We have:
Page No 99:
Question 4:
Factorize:
2a(x + y) − 3b(x + y)
Answer:
We have:
Page No 99:
Question 5:
Factorize:
2x(p2 + q2) + 4y(p2 + q2)
Answer:
We have:
Page No 99:
Question 6:
Factorize:
x(a − 5) + y(5 − a)
Answer:
We have:
Page No 99:
Question 7:
Factorize:
4(a + b) − 6(a + b)2
Answer:
We have:
Page No 99:
Question 8:
Factorize:
8(3a − 2b)2 − 10(3a − 2b)
Answer:
We have:
Page No 99:
Question 9:
Factorize:
x(x + y)3 − 3x2y(x + y)
Answer:
We have:
Page No 99:
Question 10:
Factorize:
x3 + 2x2 + 5x + 10
Answer:
We have:
Page No 99:
Question 11:
Factorize:
x2 + xy − 2xz − 2yz
Answer:
We have:
Page No 99:
Question 12:
Factorize:
a3b − a2b + 5ab − 5b
Answer:
We have:
Page No 99:
Question 13:
Factorize:
8 − 4a − 2a3 + a4
Answer:
We have:
Page No 99:
Question 14:
Factorize:
x3 − 2x2y + 3xy2 − 6y3
Answer:
We have:
Page No 99:
Question 15:
Factorize:
px − 5q + pq − 5x
Answer:
We have:
Page No 99:
Question 16:
Factorize:
x2 + y − xy − x
Answer:
We have:
Page No 99:
Question 17:
Factorize:
(3a − 1)2 − 6a + 2
Answer:
We have:
Page No 99:
Question 18:
Factorize:
(2x − 3)2 − 8x + 12
Answer:
We have:
Page No 99:
Question 19:
Factorize:
a3 + a − 3a2 − 3
Answer:
We have:
Page No 99:
Question 20:
Factorize:
3ax − 6ay − 8by + 4bx
Answer:
We have:
Page No 99:
Question 21:
Factorize:
abx2 + a2x + b2x + ab
Answer:
We have:
Page No 99:
Question 22:
Factorize:
x3 − x2 + ax + x − a − 1
Answer:
We have:
Page No 100:
Question 23:
Factorize:
2x + 4y − 8xy − 1
Answer:
We have:
Page No 100:
Question 24:
Factorize:
ab(x2 + y2) − xy(a2 + b2)
Answer:
We have:
Page No 100:
Question 25:
Factorize:
a2 + ab(b + 1) + b3
Answer:
We have:
Page No 100:
Question 26:
Factorize:
a3 + ab(1 − 2a) − 2b2
Answer:
We have:
Page No 100:
Question 27:
Factorize:
2a2 + bc − 2ab − ac2
Answer:
We have:
Page No 100:
Question 28:
Factorize:
(ax + by)2 + (bx − ay)2
Answer:
We have:
Page No 100:
Question 29:
Factorize:
a(a + b − c) − bc
Answer:
We have:
Page No 100:
Question 30:
Factorize:
a(a − 2b − c) + 2bc
Answer:
We have:
Page No 100:
Question 31:
Factorize:
a2x2 + (ax2 + 1)x + a
Answer:
We have:
Page No 100:
Question 32:
Factorize:
ab(x2 + 1) + x(a2 + b2)
Answer:
We have:
Page No 100:
Question 33:
Factorize:
x2 − (a + b)x + ab
Answer:
We have:
Page No 100:
Question 34:
Factorize:
Answer:
Page No 105:
Question 1:
Factorise:
9x2 – 16y2
Answer:
Page No 105:
Question 2:
Factorise:
Answer:
Page No 105:
Question 3:
Factorise:
81 – 16x2
Answer:
Page No 105:
Question 4:
Factorise:
5 – 20x2
Answer:
Page No 105:
Question 5:
Factorise:
2x4 – 32
Answer:
Page No 105:
Question 6:
Factorize:
3a3b − 243ab3
Answer:
Page No 105:
Question 7:
Factorize:
3x3 − 48x
Answer:
Page No 105:
Question 8:
Factorize:
27a2 − 48b2
Answer:
Page No 105:
Question 9:
Factorize:
x − 64x3
Answer:
Page No 105:
Question 10:
Factorize:
8ab2 − 18a3
Answer:
Page No 105:
Question 11:
Factorize:
150 − 6x2
Answer:
Page No 105:
Question 12:
Factorise:
2 – 50x2
Answer:
Page No 105:
Question 13:
Factorise:
20x2 – 45
Answer:
Page No 105:
Question 14:
Factorise:
(3a + 5b)2 – 4c2
Answer:
Page No 105:
Question 15:
Factorise:
a2 – b2 – a – b
Answer:
Page No 105:
Question 16:
Factorise:
4a2 – 9b2 – 2a – 3b
Answer:
Page No 105:
Question 17:
Factorise:
a2 – b2 + 2bc – c2
Answer:
Page No 105:
Question 18:
Factorise:
4a2 – 4b2 + 4a + 1
Answer:
Page No 105:
Question 19:
Factorize:
a2 + 2ab + b2 − 9c2
Answer:
Page No 105:
Question 20:
Factorize:
108a2 − 3(b − c)2
Answer:
Page No 105:
Question 21:
Factorize:
(a + b)3 − a − b
Answer:
Page No 105:
Question 22:
Factorise:
x2 + y2 – z2 – 2xy
Answer:
Page No 105:
Question 23:
Factorise:
x2 + 2xy + y2 – a2 + 2ab – b2
Answer:
Page No 105:
Question 24:
Factorise:
25x2 – 10x + 1 – 36y2
Answer:
Page No 105:
Question 25:
Factorize:
a − b − a2 + b2
Answer:
Page No 105:
Question 26:
Factorize:
a2 − b2 − 4ac + 4c2
Answer:
Page No 105:
Question 27:
Factorize:
9 − a2 + 2ab − b2
Answer:
Page No 105:
Question 28:
Factorize:
x3 − 5x2 − x + 5
Answer:
Page No 105:
Question 29:
Factorise:
1 + 2ab – (a2 + b2)
Answer:
Page No 105:
Question 30:
Factorise:
9a2 + 6a + 1 – 36b2
Answer:
Page No 105:
Question 31:
Factorize:
x2 − y2 + 6y − 9
Answer:
Page No 105:
Question 32:
Factorize:
4x2 − 9y2 − 2x − 3y
Answer:
Page No 105:
Question 33:
Factorize:
9a2 + 3a − 8b − 64b2
Answer:
Page No 105:
Question 34:
Factorise:
Answer:
Page No 105:
Question 35:
Factorise:
Answer:
Disclaimer: The expression of the question should be . The same has been done before solving the question.
Page No 105:
Question 36:
Factorise:
Answer:
Page No 105:
Question 37:
Factorise:
x8 – 1
Answer:
Page No 105:
Question 38:
Factorise:
16x4 – 1
Answer:
Page No 105:
Question 39:
81x4 – y4
Answer:
Page No 105:
Question 40:
x4 – 625
Answer:
Page No 114:
Question 1:
Factorize:
x2 + 11x + 30
Answer:
We have:
We have to split 11 into two numbers such that their sum of is 11 and their product is 30.
Clearly, .
Page No 114:
Question 2:
Factorize:
x2 + 18x + 32
Answer:
We have:
We have to split 18 into two numbers such that their sum is 18 and their product is 32.
Clearly, .
Page No 114:
Question 3:
Factorise:
x2 + 20x – 69
Answer:
Page No 114:
Question 4:
x2 + 19x – 150
Answer:
Page No 114:
Question 5:
Factorise:
x2 + 7x – 98
Answer:
Page No 114:
Question 6:
Factorise:
Answer:
Page No 114:
Question 7:
Factorise:
x2 – 21x + 90
Answer:
Page No 114:
Question 8:
Factorise:
x2 – 22x + 120
Answer:
Page No 114:
Question 9:
Factorise:
x2 – 4x + 3
Answer:
Page No 114:
Question 10:
Factorise:
Answer:
Page No 114:
Question 11:
Factorise:
Answer:
Page No 114:
Question 12:
Factorise:
Answer:
Page No 114:
Question 13:
Factorise:
Answer:
Page No 114:
Question 14:
Factorise:
Answer:
Page No 114:
Question 15:
Factorise:
x2 – 32x – 105
Answer:
Page No 114:
Question 16:
Factorise:
x2 – 11x – 80
Answer:
Page No 114:
Question 17:
Factorise:
6 – x – x2
Answer:
Page No 114:
Question 18:
Factorise:
Answer:
Page No 114:
Question 19:
Factorise:
40 + 3x – x2
Answer:
Page No 114:
Question 20:
Factorise:
x2 – 26x + 133
Answer:
Page No 114:
Question 21:
Factorise:
Answer:
Page No 114:
Question 22:
Factorise:
Answer:
Page No 114:
Question 23:
Factorise:
Answer:
Page No 114:
Question 24:
Factorise:
Answer:
Page No 114:
Question 25:
Factorize:
x2 − x − 156
Answer:
We have:
We have to split (1) into two numbers such that their sum is (1) and their product is (156).
Clearly, .
Page No 114:
Question 26:
Factorise:
x2 – 32x – 105
Answer:
Page No 114:
Question 27:
Factorise:
9x2 + 18x + 8
Answer:
Page No 114:
Question 28:
Factorise:
6x2 + 17x + 12
Answer:
Page No 114:
Question 29:
Factorize:
18x2 + 3x − 10
Answer:
We have:
We have to split 3 into two numbers such that their sum is 3 and their product is (180), i.e., .
Clearly, .
​
Page No 114:
Question 30:
Factorize:
2x2 + 11x − 21
Answer:
We have:
We have to split 11 into two numbers such that their sum is 11 and their product is (42), i.e., .
Clearly, .
Page No 114:
Question 31:
Factorize:
15x2 + 2x − 8
Answer:
We have:
We have to split 2 into two numbers such that their sum is 2 and their product is (120), i.e., .
Clearly, .
Page No 114:
Question 32:
Factorise:
21x2 + 5x – 6
Answer:
Page No 114:
Question 33:
Factorize:
24x2 − 41x + 12
Answer:
We have:
We have to split (41) into two numbers such that their sum is (41) and their product is 288, i.e., .
Clearly, .
Page No 114:
Question 34:
Factorise:
3x2 – 14x + 8
Answer:
Hence, factorisation of 3x2 – 14x + 8 is .
Page No 114:
Question 35:
Factorize:
2x2 + 3x − 90
Answer:
We have:
We have to split 3 into two numbers such that their sum is 3 and their product is (180), i.e., .
Clearly, .
Page No 114:
Question 36:
Factorize:
Answer:
We have:
We have to split 2 into two numbers such that their sum is 2 and product is (15), i.e.,.
Clearly, .
Page No 114:
Question 37:
Factorize:
Answer:
We have:
We have to split 1 into two numbers such that their sum is 1 and product is 30, i.e.,.
Clearly, .
Page No 114:
Question 38:
Factorize:
Answer:
We have:
We have to split into two numbers such that their sum is and product is 14.
Clearly, .
Page No 114:
Question 39:
Factorize:
Answer:
We have:
Now, we have to split (47) into two numbers such that their sum is (47) and their product is 90.
Clearly, .
Page No 114:
Question 40:
Factorize:
Answer:
We have:
We have to split 20 into two numbers such that their sum is 20 and their product is 75.
Clearly,
Page No 114:
Question 41:
Factorise:
Answer:
Hence, factorisation of is .
Page No 114:
Question 42:
Factorize:
Answer:
We have:
We have to split 3 into two numbers such that their sum is 3 and their product is 2, i.e., .
Clearly, .
Page No 114:
Question 43:
Factorize:
Answer:
We have:
We have to split into two numbers such that their sum is and their product is 6, i.e.,.
Clearly, .
Page No 114:
Question 44:
Factorize:
15x2 − x − 128
Answer:
We have:
We have to split (1) into two numbers such that their sum is (1) and their product is (420), i.e., .
Clearly, .
Page No 114:
Question 45:
Factorize:
6x2 − 5x − 21
Answer:
We have:
We have to split (5) into two numbers such that their sum is (5) and their product is (126), i.e., .
Clearly, .
Page No 114:
Question 46:
Factorize:
2x2 − 7x − 15
Answer:
We have:
We have to split (7) into two numbers such that their sum is (7) and their product is (30), i.e., .
Clearly, .
Page No 114:
Question 47:
Factorize:
5x2 − 16x − 21
Answer:
We have:
We have to split (16) into two numbers such that their sum is (16) and their product is (105), i.e., .
Clearly, .
Page No 114:
Question 48:
Factorise:
6x2 – 11x – 35
Answer:
Hence, factorisation of 6x2 – 11x – 35 is .
Page No 114:
Question 49:
Factorise:
9x2 – 3x – 20
Answer:
Hence, factorisation of 9x2 – 3x – 20 is .
Page No 114:
Question 50:
Factorize:
10x2 − 9x − 7
Answer:
We have:
We have to split (9) into two numbers such that their sum is (9) and their product is (70), i.e., .
Clearly, .
Page No 114:
Question 51:
Factorize:
Answer:
Now, we have to split (32) into two numbers such that their sum is (32) and their product is 112, i.e., .
Clearly, .
Page No 114:
Question 52:
Factorise:
Answer:
Hence, factorisation of is .
Page No 114:
Question 53:
Factorise:
Answer:
Hence, factorisation of is .
Page No 114:
Question 54:
Factorise:
Answer:
Hence, factorisation of is .
Page No 114:
Question 55:
Factorise:
Answer:
Hence, factorisation of is .
Page No 114:
Question 56:
Factorise:
Answer:
Hence, factorisation of is .
Page No 114:
Question 57:
Factorise:
Answer:
Hence, factorisation of is .
Page No 114:
Question 58:
Factorise:
Answer:
Hence, factorisation of is .
Page No 114:
Question 59:
Factorize:
2(x + y)2 − 9(x + y) − 5
Answer:
We have:
Thus, the given expression becomes
Now, we have to split (9) into two numbers such that their sum is (9) and their product is (10).
Clearly, .
Putting , we get:
Page No 114:
Question 60:
Factorize:
9(2a − b)2 − 4(2a − b) − 13
Answer:
We have:
Thus, the given expression becomes
Now, we must split (4) into two numbers such that their sum is (4) and their product is (117).
Clearly, .
Putting , we get:
Page No 115:
Question 61:
Factorise:
Answer:
Hence, factorisation of is .
Page No 115:
Question 62:
Factorise:
Answer:
Hence, factorisation of is .
Page No 115:
Question 63:
Factorise:
Answer:
Hence, factorisation of is .
Page No 115:
Question 64:
Factorise:
Answer:
Hence, factorisation of is .
Page No 115:
Question 65:
Factorise:
4x4 + 7x2 – 2
Answer:
Hence, factorisation of 4x4 + 7x2 – 2 is .
Page No 115:
Question 66:
Evaluate {(999)2 – 1}.
Answer:
Hence, {(999)2 – 1} = 998000.
Page No 119:
Question 1:
Expand:
(i) (a + 2b + 5c)2
(ii) (2b − b + c)2
(iii) (a − 2b − 3c)2
Answer:
Page No 119:
Question 2:
Expand:
(i) (2a − 5b − 7c)2
(ii) (−3a + 4b − 5c)2
(iii)
Answer:
Page No 119:
Question 3:
Factorize: 4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz.
Answer:
Page No 119:
Question 4:
Factorize: 9x2 + 16y2 + 4z2 − 24xy + 16yz − 12xz
Answer:
Page No 119:
Question 5:
Factorize: 25x2 + 4y2 + 9z2 − 20xy − 12yz + 30xz.
Answer:
Page No 119:
Question 6:
16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz
Answer:
Hence, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz = .
Page No 119:
Question 7:
Evaluate
(i) (99)2
(ii) (995)2
(iii) (107)2
Answer:
Page No 123:
Question 1:
Expand
(i) (3x + 2)3
(ii)
(iii)
Answer:
Page No 123:
Question 2:
Expand
(i) (5a – 3b)3
(ii)
(iii)
Answer:
Page No 123:
Question 3:
Factorise
Answer:
Hence, factorisation of is .
Page No 123:
Question 4:
Factorise
Answer:
Hence, factorisation of is .
Page No 123:
Question 5:
Factorise
Answer:
Hence, factorisation of is .
Page No 123:
Question 6:
Factorise
Answer:
Hence, factorisation of is .
Page No 123:
Question 7:
Factorise
Answer:
Hence, factorisation of is .
Page No 123:
Question 8:
Factorise
Answer:
Hence, factorisation of is .
Page No 123:
Question 9:
Factorise
a3 – 12a(a – 4) – 64
Answer:
Hence, factorisation of a3 – 12a(a – 4) – 64 is .
Page No 123:
Question 10:
Evaluate
(i) (103)3
(ii) (99)3
Answer:
Page No 129:
Question 1:
Factorize:
x3 + 27
Answer:
Page No 129:
Question 2:
Factorise
27a3 + 64b3
Answer:
We know that
Given: 27a3 + 64b3
x = 3a, y = 4b
Page No 129:
Question 3:
Factorize:
Answer:
Page No 129:
Question 4:
Factorize:
Answer:
Page No 129:
Question 5:
Factorize:
16x4 + 54x
Answer:
Page No 129:
Question 6:
Factorize:
7a3 + 56b3
Answer:
Page No 129:
Question 7:
Factorize:
x5 + x2
Answer:
Page No 129:
Question 8:
Factorize:
a3 + 0.008
Answer:
Page No 129:
Question 9:
Factorise
1 – 27a3
Answer:
Page No 129:
Question 10:
Factorize:
64a3 − 343
Answer:
Page No 129:
Question 11:
Factorize:
x3 − 512
Answer:
Page No 129:
Question 12:
Factorize:
a3 − 0.064
Answer:
Page No 129:
Question 13:
Factorize:
Answer:
Page No 129:
Question 14:
Factorise
Answer:
We know
We have,
So,
Page No 129:
Question 15:
Factorize:
x − 8xy3
Answer:
Page No 129:
Question 16:
Factorise
32x4 – 500x
Answer:
Page No 129:
Question 17:
Factorize:
3a7b − 81a4b4
Answer:
Page No 129:
Question 18:
Factorise
x4 y4 – xy
Answer:
Using the identity
Page No 129:
Question 19:
Factorise
8x2 y3 – x5
Answer:
Page No 129:
Question 20:
Factorise
1029 – 3x3
Answer:
Page No 129:
Question 21:
Factorize:
x6 − 729
Answer:
Page No 129:
Question 22:
Factorise
x9 – y9
Answer:
Page No 129:
Question 23:
Factorize:
(a + b)3 − (a − b)3
Answer:
Page No 129:
Question 24:
Factorize:
8a3 − b3 − 4ax + 2bx
Answer:
Page No 129:
Question 25:
Factorize:
a3 + 3a2b + 3ab2 + b3 − 8
Answer:
Page No 129:
Question 26:
Factorize:
Answer:
Page No 129:
Question 27:
Factorize:
2a3 + 16b3 − 5a − 10b
Answer:
Page No 129:
Question 28:
Factorise
a6 + b6
Answer:
Page No 129:
Question 29:
Factorise
a12 – b12
Answer:
a12 – b12
Page No 129:
Question 30:
Factorise
x6 – 7x3 – 8
Answer:
Let
So, the equation becomes
Page No 129:
Question 31:
Factorise
x3 – 3x2 + 3x + 7
Answer:
x3 – 3x2 + 3x + 7
Page No 129:
Question 32:
Factorise
(x +1)3 + (x – 1)3
Answer:
(x +1)3 + (x – 1)3
Page No 129:
Question 33:
Factorise
(2a +1)3 + (a – 1)3
Answer:
(2a +1)3 + (a – 1)3
Page No 129:
Question 34:
Factorise
8(x +y)3 – 27(x – y)3
Answer:
8(x +y)3 – 27(x – y)3
Page No 129:
Question 35:
Factorise
(x +2)3 + (x – 2)3
Answer:
(x +2)3 + (x – 2)3
Page No 129:
Question 36:
Factorise
(x + 2)3 – (x – 2)3
Answer:
(x + 2)3 – (x – 2)3
Page No 129:
Question 37:
Prove that .
Answer:
Thus, LHS = RHS
Page No 129:
Question 38:
Prove that .
Answer:
Thus, LHS=RHS
Page No 136:
Question 1:
Find the product:
(x + y − z) (x2 + y2 + z2 − xy + yz + zx)
Answer:
Page No 136:
Question 2:
Find the product:
(x – y − z) (x2 + y2 + z2 + xy – yz + xz)
Answer:
(x – y − z) (x2 + y2 + z2 + xy – yz + xz)
Page No 136:
Question 3:
Find the product:
(x − 2y + 3) (x2 + 4y2 + 2xy − 3x + 6y + 9)
Answer:
Page No 136:
Question 4:
Find the product:
(3x – 5y + 4) (9x2 + 25y2 + 15xy − 20y + 12x + 16)
Answer:
Page No 136:
Question 5:
Factorize:
125a3 + b3 + 64c3 − 60abc
Answer:
Page No 136:
Question 6:
Factorize:
a3 + 8b3 + 64c3 − 24abc
Answer:
Page No 136:
Question 7:
Factorize:
1 + b3 + 8c3 − 6bc
Answer:
Page No 136:
Question 8:
Factorize:
216 + 27b3 + 8c3 − 108abc
Answer:
Page No 136:
Question 9:
Factorize:
27a3 − b3 + 8c3 + 18abc
Answer:
Page No 136:
Question 10:
Factorize:
8a3 + 125b3 − 64c3 + 120abc
Answer:
Page No 136:
Question 11:
Factorize:
8 − 27b3 − 343c3 − 126bc
Answer:
Page No 136:
Question 12:
Factorize:
125 − 8x3 − 27y3 − 90xy
Answer:
Page No 137:
Question 13:
Factorize:
Answer:
Page No 137:
Question 14:
Factorise:
27x3 – y3 – z3 – 9xyz
Answer:
Page No 137:
Question 15:
Factorise:
Answer:
Page No 137:
Question 16:
Factorise:
Answer:
Page No 137:
Question 17:
Factorize:
(a − b)3 + (b − c)3 + (c − a)3
Answer:
Page No 137:
Question 18:
Factorise:
Answer:
Page No 137:
Question 19:
Factorize:
(3a − 2b)3 + (2b − 5c)3 + (5c − 3a)3
Answer:
Page No 137:
Question 20:
Factorize:
(5a − 7b)3 + (9c − 5a)3 + (7b − 9c)3
Answer:
Page No 137:
Question 21:
Factorize:
a3(b − c)3 + b3(c − a)3 + c3(a − b)3
Answer:
Page No 137:
Question 22:
Evaluate
(i) (–12)3 + 73 + 53
(ii) (28)3 + (–15)3 + (–13)3
Answer:
(i) (–12)3 + 73 + 53
(ii) (28)3 + (–15)3 + (–13)3
Page No 137:
Question 23:
Prove that
Answer:
Page No 137:
Question 24:
If a, b, c are all nonzero and a + b + c = 0, prove that .
Answer:
Thus, we have:
Page No 137:
Question 25:
If a + b + c = 9 and a2 + b2 + c2 = 35, find the value of (a3 + b3 + c3 – 3abc).
Answer:
a + b + c = 9
We know,
(a3 + b3 + c3 – 3abc) =
Page No 138:
Question 1:
If (x + 1) is factor of the polynomial (2x2 + kx) then the value of k is
(a) –2
(b) –3
(c) 2
(d) 3
Answer:
(c) 2
Page No 138:
Question 2:
The value of (249)2 – (248)2 is
(a) 12
(b) 477
(c) 487
(d) 497
Answer:
(249)2 – (248)2
We know
Hence, the correct answer is option (d).
Page No 138:
Question 3:
If , where x ≠ 0 and y ≠ 0, then the value of (x3 − y3) is
(a) 1
(b) −1
(c) 0
(d)
Answer:
(c) 0
Thus, we have:
Page No 138:
Question 4:
If a + b + c = 0, then a3 + b3 + c3 = ?
(a) 0
(b) abc
(c) 2abc
(d) 3abc
Answer:
(d) 3abc
Page No 139:
Question 5:
If then the value of p is
(a) 0
(b)
(c)
(d)
Answer:
Hence, the correct answer is option (c).
Page No 139:
Question 6:
The coefficient of x in the expansion of (x + 3)3 is
(a) 1
(b) 9
(c) 18
(d) 27
Answer:
(x + 3)3
So, the coefficient of x in (x + 3)3 is 27.
Hence, the correct answer is option (d).
Page No 139:
Question 7:
Which of the following is a factor of (x + y)3 – (x3 + y3)?
(a) x2 + y2 + 2xy
(b) x2 + y2 – xy
(c) xy2
(d) 3xy
Answer:
(x + y)3 – (x3 + y3)
Thus, the factors of (x + y)3 – (x3 + y3) are 3xy and (x + y).
Hence, the correct answer is option (d).
Page No 139:
Question 8:
One of the factors of is
(a) 5 + x
(b) 5 – x
(c) 5x – 1
(d) 10x
Answer:
So, the factors of are (5x + 1) and 10x
Hence, the correct answer is option (d).
Page No 139:
Question 9:
If (x + 5) is a factor of p(x) = x3 − 20x + 5k, then k = ?
(a) −5
(b) 5
(c) 3
(d) −3
Answer:
(b) 5
Page No 139:
Question 10:
If (x + 2) and (x − 1) are factors of (x3 + 10x2 + mx + n), then
(a) m = 5, n = −3
(b) m = 7, n = −18
(c) m = 17, n = −8
(d) m = 23, n = −19
Answer:
(b) m = 7, n = −18
Let:
Now,
(x + 2) is a factor of p(x).
So, we have p(2)=0
Now,
Also,
(x 1) is a factor of p(x).
We have:
p(1) = 0
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18
Page No 139:
Question 11:
104 × 96 = ?
(a) 9864
(b) 9984
(c) 9684
(d) 9884
Answer:
(b) 9984
Page No 139:
Question 12:
305 × 308 = ?
(a) 94940
(b) 93840
(c) 93940
(d) 94840
Answer:
(c) 93940
Page No 139:
Question 13:
207 × 193 = ?
(a) 39851
(b) 39951
(c) 39961
(d) 38951
Answer:
(b) 39951
Page No 139:
Question 14:
4a2 + b2 + 4ab + 8a + 4b + 4 = ?
(a) (2a + b + 2)2
(b) (2a − b + 2)2
(c) (a + 2b + 2)2
(d) none of these
Answer:
(a) (2a + b + 2)2
Page No 139:
Question 15:
(x2 − 4x − 21) = ?
(a) (x − 7)(x − 3)
(b) (x + 7)(x − 3)
(c) (x − 7)(x + 3)
(d) none of these
Answer:
(c) (x − 7)(x + 3)
Page No 139:
Question 16:
(4x2 + 4x − 3) = ?
(a) (2x − 1) (2x − 3)
(b) (2x + 1) (2x − 3)
(c) (2x + 3) (2x − 1)
(d) none of these
Answer:
(c) (2x + 3) (2x − 1)
Page No 139:
Question 17:
6x2 + 17x + 5 = ?
(a) (2x + 1)(3x + 5)
(b) (2x + 5)(3x + 1)
(c) (6x + 5)(x + 1)
(d) none of these
Answer:
(b) (2x + 5)(3x + 1)
Page No 139:
Question 18:
(x + 1) is a factor of the polynomial
(a) x3 − 2x2 + x + 2
(b) x3 + 2x2 + x − 2
(c) x3 − 2x2 − x − 2
(d) x3 − 2x2 − x + 2
Answer:
(c) x3 − 2x2 − x − 2
Let:
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:
Hence, (x + 1) is not a factor of .
Now,
Let:
By the factor theorem, (x + 1) will be a factor of f (x) if f (1) = 0.
We have:
Hence, (x + 1) is not a factor of .
Now,
Let:
By the factor theorem, (x + 1) will be a factor of f (x) if f (1) = 0.
We have:
Hence, (x + 1) is a factor of .
Page No 140:
Question 19:
3x3 + 2x2 + 3x + 2 = ?
(a) (3x − 2)(x2 − 1)
(b) (3x − 2)(x2 + 1)
(c) (3x + 2)(x2 − 1)
(d) (3x + 2)(x2 + 1)
Answer:
(d) (3x + 2)(x2 + 1)
Page No 140:
Question 20:
If a + b + c = 0, then =?
Answer:
(d) 3
Thus, we have:
Page No 140:
Question 21:
If x + y + z = 9 and xy + yz + zx = 23, then the value of (x3 + y3 + z3 − 3xyz) = ?
(a) 108
(b) 207
(c) 669
(d) 729
Answer:
(a) 108
Page No 140:
Question 22:
If then (a3 − b3) = ?
(a) −3
(b) −2
(c) −1
(d) 0
Answer:
2 + b2 = ab
2 + b2 + ab = 0
Thus, we have:
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