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Exercise 5(A)
Factorise by taking out the common factors:
1. 2(2x – 5y) (3x + 4y) – 6(2x – 5y) (x – y)
Solution:
Identifying and taking (2x – 5y) common from both the terms, we have
= (2x – 5y) [2(3x + 4y) – 6(x – y)]
= (2x – 5y) (6x + 8y – 6x + 6y)
= (2x – 5y) (8y + 6y)
= (2x – 5y) (14y)
= (2x – 5y)14y
2. xy(3x2 – 2y2) – yz(2y2 – 3x2) + zx(15x2 – 10y2)
Solution:
We have, xy(3x2 – 2y2) – yz(2y2 – 3x2) + zx(15x2 – 10y2)
Changing signs to arrive at a common term
So,
= xy(3x2 – 2y2) + yz(3x2 – 2y2) + zx(15x2 – 10y2)
= xy(3x2 – 2y2) + yz(3x2 – 2y2) + 5zx(3x2 – 2y2)
= (3x2 – 2y2) (xy + yz + 5zx)
3. ab(a2 + b2 – c2) – bc(c2 – a2 – b2) + ca(a2 + b2 – c2)
Solution:
We have, ab(a2 + b2 – c2) – bc(c2 – a2 – b2) + ca(a2 + b2 – c2)
Changing signs to arrive at a common term
So,
= ab(a2 + b2 – c2) + bc(a2 + b2 – c2) + ca(a2 + b2 – c2)
= (a2 + b2 – c2) (ab + bc + ca)
4. 2x(a – b) + 3y(5a – 5b) + 4z(2b – 2a)
Solution:
We have, 2x(a – b) + 3y(5a – 5b) + 4z(2b – 2a)
Taking common factors, we get
= 2x(a – b) + 15y(a – b) – 8z(a – b)
= (a – b) (2x + 15y – 8z)
Factorize by the grouping method:
5. a3 + a – 3a2 – 3
Solution:
We have, a3 + a – 3a2 – 3
Grouping to arrive at a common term
= a (a2 + 1) – 3(a2 + 1)
Taking common, we get
= (a2 + 1) (a -3)
6. 16(a + b)2 – 4a – 4b
Solution:
We have, 16 (a + b)2 – 4a – 4b
Grouping to arrive at a common term
= 16(a + b)2 – 4 (a + b)
Taking common, we get
= 4(a + b) [4 (a + b) – 1]
= 4(a + b) (4a + 4b – 1)
7. Factorize by the grouping method:
a4 – 2a3 – 4a + 8
Solution:
We have, a4 – 2a3 – 4a + 8
Grouping to arrive at a common term
= a3(a – 2) – 4(a – 2)
Taking common, we get
= (a3 – 4) (a – 2)
8. ab – 2b + a2 – 2a
Solution:
We have, ab – 2b + a2 – 2a
Grouping to arrive at a common term
= b(a – 2) + a(a – 2)
Taking common, we get
= (b + a) (a – 2)
9. ab(x2 + 1) + x(a2 + b2)
Solution:
We have, ab(x2 + 1) + x(a2 + b2)
On expanding,
= abx2 + ab + a2x + b2x
Now, grouping to arrive at a common term
= abx2 + a2x + b2x + ab
= ax(bx + a) + b(bx + a)
Taking common, we get
= (ax + b) (bx + a)
10. a2 + b – ab – a
Solution:
We have, a2 + b – ab – a
Grouping to arrive at a common term
= a2 – a + b – ab
= a(a – 1) – b(-1 + a)
= a(a – 1) – b(a – 1)
Taking common, we get
= (a – b) (a – 1)
11. (ax + by)2 + (bx – ay)2
Solution:
We have, (ax + by)2 + (bx – ay)2
On expanding,
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2abxy
= a2x2 + b2y2 + b2x2 + a2y2
Rearranging terms, we get
= a2x2 + b2x2 + a2y2 + b2y2
Taking common, we get
= x2(a2 + b2) + y2(a2 + b2)
= (x2 + y2) (a2 + b2)
12. a2x2 + (ax2 + 1)x + a
Solution:
We have, a2x2 + (ax2 + 1)x + a
Regrouping the terms, we have
= a2x2 + a + (ax2 + 1)x
= a(ax2 + 1) + x(ax2 + 1)
Taking common, we get
= (ax2 + 1) (a + x)
13. (2a – b)2 – 10a + 5b
Solution:
We have, (2a – b)2 – 10a + 5b
Taking common,
= (2a – b)2 – 5(2a – b)
Now,
= (2a – b) [(2a – b) – 5]
= (2a – b) (2a – b – 5)
14. a(a – 4) – a + 4
Solution:
We have, a(a – 4) – a + 4
By grouping, we get
= a(a – 4) -1(a – 4)
Now, taking the common term
= (a – 4) (a – 1)
15. y2 – (a + b) y + ab
Solution:
We have, y2 – (a + b) y + ab
On expanding,
= y2 – ay – by + ab
= (y2 – ay) – by + ab
Taking ‘y’ and ‘b’ common from the group, we get
= y(y – a) – b(y – a)
= (y – a) (y – b)
16. a2 + 1/a2 – 2 – 3a + 3/a
Solution:
We have, a2 + 1/a2 – 2 – 3a + 3/a
On grouping terms, we get
= (a2 – 2 + 1/a2) – 3a + 3/a
= [a2 – (2 x a x 1/a) + 1/a2] – 3(a – 1/a)
= (a – 1/a)2 – 3(a – 1/a) {Since, (x – y)2 = x2 – 2xy + y2}
Taking (a – 1/a) as common, we get
= (a – 1/a) [(a – 1/a) – 3]
= (a – 1/a) (a – 1/a – 3)
17. x2 + y2 + x + y + 2xy
Solution:
We have, x2 + y2 + x + y + 2xy
On rearranging terms, we get
= (x2 + y2 + 2xy) + (x + y) {Since, (x + y)2 = x2 + 2xy + y2}
Now,
= (x + y)2 + (x + y)
= (x + y)(x + y + 1)
18. a2 + 4b2 – 3a + 6b – 4ab
Solution:
We have, a2 + 4b2 – 3a + 6b – 4ab
On rearranging terms, we get
= a2 + 4b2 – 4ab – 3a + 6b
Now,
= a2 + (2b)2 – 2 × a × (2b) – 3(a – 2b) {Since, (a – b)2 = a2 – 2ab + b2}
= (a – 2b)2 – 3(a – 2b)
= (a – 2b) [(a – 2b)- 3]
= (a – 2b) (a – 2b – 3)
19. m (x – 3y)2 + n (3y – x) + 5x – 15y
Solution:
We have, m (x – 3y)2 + n (3y – x) + 5x – 15y
Now,
Taking (x – 3y) common from all the three terms, we get
= m(x – 3y)2 – n(x – 3y) + 5(x – 3y)
= (x – 3y) [m(x – 3y) – n + 5]
= (x – 3y) (mx – 3my – n + 5)
20. x(6x – 5y) – 4(6x – 5y)2
Solution:
We have, x(6x – 5y) – 4(6x – 5y)2
Now,
Taking (6x – 5y) common from the three terms, we get
= (6x – 5y) [x – 4(6x – 5y)]
= (6x – 5y) (x – 24x + 20y)
= (6x – 5y) (-23x + 20y)
= (6x – 5y) (20y – 23x)
Exercise 5(B)
Factorize:
1. a2 + 10a + 24
Solution:
We have, a2 + 10a + 24
By splitting the middle term, we get
= a2 + 6a + 4a + 24
= a(a + 6) + 4(a + 6)
= (a + 4) (a + 6)
2. a2 – 3a – 40
Solution:
We have, a2 – 3a – 40
By splitting the middle term, we get
= a2 – 8a + 5a – 40
= a(a – 8) + 5(a – 8)
= (a + 5) (a – 8)
3. 1 – 2a – 3a2
Solution:
We have, 1 – 2a – 3a2
By splitting the middle term, we get
= 1 – 3a + a – 3a2
= 1(1 – 3a) + a(1 – 3a)
= (1 + a) (1 – 3a)
4. x2 – 3ax – 88a2
Solution:
We have, x2 – 3ax – 88a2
By splitting the middle term, we get
= x2 – 11ax + 8ax – 88a2
= x(x – 11a) + 8a(x – 11a)
= (x + 8a) (x – 11a)
5. 6a2 – a – 15
Solution:
We have, 6a2 – a – 15
By splitting the middle term, we get
= 6a2 + 9a – 10a – 15
= 3a(2a + 3) – 5(2a + 3)
= (3a – 5) (2a + 3)
6. 24a3 + 37a2 – 5a
Solution:
We have, 24a3 + 37a2 – 5a
Taking ‘a’ common from all
= a (24a2 + 37a – 5)
= a (24a2 + 40a – 3a – 5) {By splitting the middle term}
= a [8a(3a + 5) – 1(3a + 5)]
= a [(8a – 1) (3a + 5)]
= a (8a – 1) (3a + 5)
7. a(3a – 2) – 1
Solution:
We have, a(3a – 2) – 1
On expanding,
= 3a2 – 2a – 1
By splitting the middle term, we get
= 3a2 – 3a + a – 1
= 3a(a – 1) + 1(a – 1)
= (3a + 1) (a – 1)
8. a2b2 + 8ab – 9
Solution:
We have, a2b2 + 8ab – 9
By splitting the middle term, we get
= a2b2 + 9ab – ab – 9
= ab(ab + 9) – 1(ab + 9)
= (ab – 1) (ab + 9)
9. 3 – a (4 + 7a)
Solution:
We have, 3 – a (4 + 7a)
On expanding,
= 3 – 4a – 7a2
By splitting the middle term, we get
= 3 + 3a – 7a – 7a2
= 3(1 + a) – 7a(1 + a)
= (1 + a) (3 – 7a)
10. (2a + b)2 – 6a – 3b – 4
Solution:
We have, (2a + b)2 – 6a – 3b – 4
= (2a + b)2 – 3(2a + b) – 4
Let’s assume that (2a + b) = x
So, the expression becomes
= x2 – 3x – 4
By splitting the middle term, we get
= x2 – 4x + x – 4
= x(x – 4) + 1(x – 4)
= (x – 4) (x + 1)
Resubstituting the value of x, we get
= (2a + b – 4) (2a + b + 1)
11. 1 – 2 (a+ b) – 3 (a + b)2
Solution:
We have, 1 – 2 (a+ b) – 3 (a + b)2
Let’s assume (a + b) = x
Then, the expression becomes
= 1 – 2x – 3x2
By splitting the middle term, we get
= 1 – 3x + x – 3x2
= 1(1 – 3x) + x(1 – 3x)
= (1 – 3x) (1 + x)
Resubstituting the value of x, we get
= [1 – 3(a + b)] [1 + (a + b)]
= (1 – 3a – 3b) (1 + a + b)
12. 3a2 – 1 – 2a
Solution:
We have, 3a2 – 1 – 2a
Rearranging,
= 3a2 – 2a – 1
By splitting the middle term, we get
= 3a2 – 3a + a – 1
= 3a(a – 1) + 1(a – 1)
= (3a + 1) (a – 1)
13. x2 + 3x + 2 + ax + 2a
Solution:
We have, x2 + 3x + 2 + ax + 2a
By splitting the middle term, we get
= (x2 + 2x + x + 2) + ax + 2a
= x(x + 2) + 1(x + 2) + a(x + 2)
= (x + 2) (x + a + 1)
14. (3x – 2y)2 + 3 (3x – 2y) – 10
Solution:
We know, (3x – 2y)2 + 3 (3x – 2y) – 10
Let’s assume that (3x – 2y) = a
So, the expression becomes
= a2 + 3a – 10
By splitting the middle term, we get
= a2 + 5a – 2a – 10
= a(a + 5) – 2(a + 5)
= (a – 2) (a + 5)
= (3x – 2y + 5) (3x – 2y – 2)
15. 5 – (3a2 – 2a) (6 – 3a2 + 2a)
Solution:
Given, 5 – (3a2 – 2a) (6 – 3a2 + 2a)
= 5 – (3a2 – 2a) [6 – (3a2 – 2a)]
Let’s substitute (3a2 – 2a) = x
And, the expression becomes,
= 5 – x(6 – x)
= 5 – 6x – x2
= 5 – 5x – x – x2
= 5(1 – x) – x(1 – x)
= (1 – x) (5 – x)
= (x – 1) (x – 5)
= (3a2 – 2a – 1) (3a2 – 2a – 5)
Now,
= (3a2 – 3a + a – 1) (3a2 + 3a – 5a – 5) {By splitting the middle term}
= [3a(a – 1) + 1(a – 1)] [3a(a + 1) – 5(a + 1)]
= [(3a + 1) (a – 1)] [(3a – 5) (a + 1)]
= (3a + 1) (3a – 5) (a + 1)(a – 1)
16. 1/35 + 12a/35 + a2
Solution:
We have, 1/35 + 12a/35 + a2
Taking common,
= 1/35 (1 + 12a + 35a2)
= 1/35 (35a2 + 12a + 1)
= 1/35 (35a2 + 7a + 5a + 1) {By splitting the middle term}
= 1/35 [7a(5a + 1) + 1(5a + 1)]
= 1/35 [(7a + 1) (5a + 1)]
= [(7a + 1) (5a + 1)]/ 35
17. (x2 – 3x) (x2 – 3x – 1) – 20.
Solution:
We have, (x2 – 3x) (x2 – 3x – 1) – 20
= (x2 – 3x)[(x2 – 3x) – 1] – 20
Let’s
= a[a – 1] – 20 ….(Taking x2 – 3x = a)
= a2 – a – 20
= a2 – 5a + 4a – 20
= a(a – 5) + 4(a – 5)
= (a – 5)(a + 4)
= (x2 – 3x – 5)(x2 – 3x + 4)
18. Find each trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
(i) x2 – 3x – 54
(ii) 2x2 – 7x – 15
(iii) 2x2 + 2x – 75
(iv) 3x2 + 4x – 10
(v) x(2x – 1) – 1
Solution:
(i) Given, x2 – 3x – 54
On comparing with the general form ax2 + bx + c, we get
a = 1, b = -3 and c = -54
So, b2 – 4ac = (-3)2 – 4(1)(-54) = 9 + 216 = 225
225 is a perfect square
Thus, x2 – 3x – 54 is factorisable
Now,
x2 – 3x – 54 = x2 – 9x + 6x – 54
= x(x – 9) + 6(x – 9)
= (x + 6) (x – 9)
(ii) Given, 2x2 – 7x – 15
On comparing with the general form ax2 + bx + c, we get
a = 2, b = -7 and c = -15
So, b2 – 4ac = (-7)2 – 4(2)(-15) = 49 + 120 = 169
169 is a perfect square
Thus, 2x2 – 7x – 15 is factorisable
Now,
2x2 – 7x – 15 = 2x2 – 10x + 3x – 15
= 2x(x – 5) + 3(x – 5)
= (2x + 3) (x – 5)
(iii) Given, 2x2 + 2x – 75
On comparing with the general form ax2 + bx + c, we get
a = 2, b = 2 and c = -75
So, b2 – 4ac = (2)2 – 4(2)(-75) = 4 + 600 = 604
604 is not a perfect square
Thus, 2x2 + 2x – 75 is not factorizable
(iv) Given, 3x2 + 4x – 10
On comparing with the general form ax2 + bx + c, we get
a = 3, b = 4 and c = -10
So, b2 – 4ac = (4)2 – 4(3)(-10) = 16 + 120 = 136
136 is a not perfect square
Thus, 3x2 + 4x – 10 is not factorizable
(v) Given, x(2x – 1) – 1
= 2x2 – x – 1
On comparing with the general form ax2 + bx + c, we get
a = 2, b = -1 and c = -1
So, b2 – 4ac = (-1)2 – 4(2)(-1) = 1 + 8 = 9
9 is a perfect square
Thus, x(2x – 1) – 1 is factorisable
Now,
x(2x – 1) – 1 = 2x2 – x – 1
= 2x2 – 2x + x – 1
= 2x(x – 1) + 1(x – 1)
= (2x + 1) (x – 1)
19. Factorise:
(i) 4√3x2 + 5x – 2√3
(ii) 7√2x2 – 10x – 4√2
Solution:
(i) We have, 4√3x2 + 5x – 2√3
By splitting the middle term, we get
= 4√3x2 + 8x – 3x – 2√3
= 4x(√3x + 2) – √3(√3x + 2)
= (4x – √3) (√3x + 2)
(ii) We have, 7√2x2 – 10x – 4√2
By splitting the middle term, we get
= 7√2x2 – 14x + 4x – 4√2
= 7√2x(x – √2) + 4(x – √2)
= (7√2x + 4) (x – √2)
20. Give possible expressions for the length and the breadth of the rectangle whose area is 12x2 – 35x + 25.
Solution:
We have, 12x2 – 35x + 25
By splitting the middle term, we get
= 12x2 – 20x – 15x + 25
= 4x(3x – 5) – 5(3x – 5)
= (3x – 5) (4x – 5)
Hence,
Length = (3x – 5) and breadth = (4x – 5) or,
Length = (4x – 5) and breadth = (3x – 5)
Exercise 5(C)
Factorize:
1. 25a2 – 9b2
Solution:
We have, 25a2 – 9b2
= (5a)2 – (3b)2
= (5a + 3b) (5a – 3b) [As x2 – y2 = (x + y)(x – y)]
2. a2 – (2a + 3b)2
Solution:
We have, a2 – (2a + 3b)2
= [a – (2a + 3b)] [a + (2a + 3b)] [As x2 – y2 = (x + y)(x – y)]
= (a – 2a – 3b) (a + 2a + 3b)
= (-a – 3b) (3a + 3b)
= -3(a + 3b) (a + b)
3. a2 – 81(b-c)2
Solution:
We have, a2 – 81(b-c)2
= a2 – [9(b – c)]2
= [a – 9(b – c)] [a + 9(b – c)]] [As x2 – y2 = (x + y)(x – y)]
= (a – 9b + 9c) (a + 9b – 9c)
4. 25(2a – b)2 – 81b2
Solution:
We have, 25(2a – b)2 – 81b2
= [5(2a – b)]2 – (9b)2
= [5(2a – b) – 9b] [5(2a – b) + 9b] [As x2 – y2 = (x + y)(x – y)]
= (10a – 5b – 9b) (10a – 5b + 9b)
= (10a – 14b) (10a + 4b)
= 2(5a – 7b). 2(5a + 2b)
= 2(5a – 7b) (5a + 2b)
5. 50a3 – 2a
Solution:
We have, 50a3 – 2a
= 2a(25a2 – 1)
= 2a[(5a)2 – 12]
= 2a(5a – 1)(5a + 1) [As x2 – y2 = (x + y)(x – y)]
6. 4a2b – 9b3
Solution:
We have, 4a2b – 9b3
= b(4a2 – 9b2)
= b[(2a)2 – (3b)2]
= b[(2a + 3b) (2a – 3b)] [As x2 – y2 = (x + y)(x – y)]
= b(2a + 3b)(2a – 3b)
7. 3a5 – 108a3
Solution:
We have, 3a5 – 108a3
= 3a3(a2 – 36)
= 3a3(a2 – 62)
= 3a3(a – 6) (a + 6) [As x2 – y2 = (x + y)(x – y)]
8. 9(a – 2)2 – 16(a + 2)2
Solution:
We have, 9(a – 2)2 – 16(a + 2)2
= [3(a – 2)]2 – [4(a + 2)]2
= [3(a – 2) – 4(a + 2)] [3(a – 2) + 4(a + 2)] [As x2 – y2 = (x + y)(x – y)]
= [3a – 6 – 4a – 8] [3a – 6 + 4a + 8]
= [-a – 14] [7a + 2]
= -(a + 14) (7a + 2)
9. a4 – 1
Solution:
We have, a4 – 1
= (a2)2 – 12
= (a2 – 1) (a2 + 1) [As x2 – y2 = (x + y)(x – y)]
= [(a – 1)(a + 1)] (a2 + 1)
= (a – 1)(a + 1)(a2 + 1)
10. a3 + 2a2 – a – 2
Solution:
We have, a3 + 2a2 – a – 2
= a2(a + 2) – 1(a + 2)
= (a2 – 1) (a + 2)
= (a – 1) (a + 1) (a + 2) [As x2 – y2 = (x + y)(x – y)]
11. (a + b)3 – a – b
Solution:
We have, (a + b)3 – a – b
= (a + b)3 – (a + b)
= (a + b) [(a + b)2 – 1]
= (a + b) [(a + b – 1) (a + b + 1)] [As x2 – y2 = (x + y)(x – y)]
= (a + b) (a + b -1) (a + b + 1)
12. a(a – 1) – b(b – 1)
Solution:
We have, a(a – 1) – b(b – 1)
= a2 – a – b2 + b
= (a2 – b2) – (a – b)
= (a + b) (a – b) – (a – b) [As x2 – y2 = (x + y)(x – y)]
= (a – b) [(a + b) – 1]
= (a – b) (a + b – 1)
13. 4a2 – (4b2 + 4bc + c2)
Solution:
We know, 4a2 – (4b2 + 4bc + c2)
= (2a)2 – [(2b)2 + 2(2b)(c) + c2]
= (2a)2 – (2b + c)2
= (2a – 2b – c) (2a + 2b + c) [As x2 – y2 = (x + y)(x – y)]
14. 4a2 – 49b2 + 2a – 7b
Solution:
We know, 4a2 – 49b2 + 2a – 7b
= (2a)2 – (7b)2 + (2a – 7b)
= [(2a – 7b) (2a + 7b)] + (2a – 7b) [As x2 – y2 = (x + y)(x – y)]
= (2a – 7b) [(2a + 7b) + 1]
= (2a – 7b) (2a + 7b + 1)
15. 9a2 + 3a – 8b – 64b2
Solution:
We have, 9a2 + 3a – 8b – 64b2
= 9a2 – 64b2 + 3a – 8b
= (3a)2 – (8b)2 + (3a – 8b)
= [(3a – 8b) (3a + 8b)] + (3a – 8b) [As x2 – y2 = (x + y)(x – y)]
= (3a – 8b) [(3a + 8b) + 1]
= (3a – 8b) (3a + 8b + 1)
16. 4a2 – 12a + 9 – 49b2
Solution:
We have, 4a2 – 12a + 9 – 49b2
= [(2a)2 – 2(2a)(3) + 32] – (7b)2
= (2a – 3)2 – (7b)2
= (2a – 7b – 3) (2a + 7b – 3) [As x2 – y2 = (x + y)(x – y)]
17. 4xy – x2 – 4y2 + z2
Solution:
We have, 4xy – x2 – 4y2 + z2
On rearranging,
= z2 – x2 – 4y2 + 4xy
= z2 – (x2 + 4y2 – 4xy)
= z2 – (x – 2y)2
= (z – x + 2y) (z + x – 2y) [As x2 – y2 = (x + y)(x – y)]
18. a2 + b2 – c2 – d2 + 2ab – 2cd
Solution:
We have, a2 + b2 – c2 – d2 + 2ab – 2cd
On rearranging,
= a2 + 2ab + b2 – c2 – d2 – 2cd
= (a2 + 2ab + b2) – (c2 + d2 + 2cd)
= (a + b)2 – (c + d)2
= (a + b + c + d) (a + b – c – d) [As x2 – y2 = (x + y)(x – y)]
19. 4x2 – 12ax – y2 – z2 – 2yz + 9a2
Solution:
We have, 4x2 – 12ax – y2 – z2 – 2yz + 9a2
On rearranging,
= 4x2 – 12ax + 9a2 – y2 – z2 – 2yz
= (4x2 – 12ax + 9a2) – (y2 + z2 + 2yz)
= (2x – 3a)2 – (y + z)2
= (2x – 3a + y + z) (2x – 3a – y – z) [As x2 – y2 = (x + y)(x – y)]
20. (a2 – 1) (b2 – 1) + 4ab
Solution:
We have, (a2 – 1) (b2 – 1) + 4ab
By cross multiplying and expanding, we get
= (1 – a2 – b2 + a2b2) + 4ab
On manipulating,
= (a2b2 + 1 + 2ab) – (a2 + b2 – 2ab)
Now,
= (ab + 1)2 – (a – b)2
= [(ab + 1) – (a – b)] [(ab + 1) + (a – b)]
= (ab + 1 – a + b) (ab + 1 + a – b) [As x2 – y2 = (x + y)(x – y)]
21. x4 + x2 + 1
Solution:
We have, x4 + x2 + 1
= x4 + 2x2 + 1 – x2
= (x2 + 1)2 – x2 [As a2 – b2 = (a + b)(a – b)]
= (x2 + 1 – x) (x2 + 1 + x)
22. (a2 + b2 – 4c2)2 – 4a2b2
Solution:
We have, (a2 + b2 – 4c2)2 – 4a2b2
= (a2 + b2 – 4c2)2 – (2ab)2
= [(a2 + b2 – 4c2) + (2ab)] [(a2 + b2 – 4c2) – (2ab)] [As x2 – y2 = (x + y)(x – y)]
= [(a2 + b2 + 2ab) – 4c2] [(a2 + b2 – 2ab) – 4c2]
= [(a + b)2 – (2c)2] [(a – b)2 – (2c)2]
= [(a + b – 2c) (a + b + 2c)] [(a – b – 2c) (a – b + 2c)] [As x2 – y2 = (x + y)(x – y)]
= (a + b – 2c) (a + b + 2c) (a – b – 2c) (a – b + 2c)
23. (x2 + 4y2 – 9z2)2 – 16x2y2
Solution:
We have, (x2 + 4y2 – 9z2)2 – 16x2y2
= (x2 + 4y2 – 9z2)2 – (4xy)2
= (x2 + 4y2 – 9z2 – 4xy) (x2 + 4y2 – 9z2 + 4xy) [As x2 – y2 = (x + y)(x – y)]
= [(x2 – 4xy + 4y2) – 9z2] [(x2 + 4xy + 4y2) – 9z2]
= [(x – 2y)2 – (3z)2] [(x + 2y)2 – (3z)2]
= [(x – 2y + 3z) (x – 2y – 3z)] [(x +2y + 3z) (x + 2y – 3z)] [As x2 – y2 = (x + y)(x – y)]
= (x – 2y + 3z) (x – 2y – 3z)] [(x +2y + 3z) (x + 2y – 3z)
24. (a + b) 2 – a2 + b2
Solution:
We have, (a + b) 2 – a2 + b2
On expanding,
= (a2 + 2ab + b2) – a2 + b2
= 2b2 + 2ab
= 2b (b + a)
25. a2 – b2 – (a + b) 2
Solution:
We have, a2 – b2 – (a + b)
On expanding,
= a2 – b2 – (a2 + b2 + 2ab)
= a2 – b2 – a2 – b2 – 2ab
= -2b2 – 2ab
= -2b(b + a)
26. 9a2 – (a2 – 4)2
Solution:
We have, 9a2 – (a2 – 4)2
= (3a)2 – (a2 – 4)2
= [3a – (a2 – 4)] [3a + (a2 – 4)] [As x2 – y2 = (x + y)(x – y)]
= [3a – (a2 – 22)] [3a + (a2 – 22)]
= (3a – a2 + 4) (3a + a2 – 4)
= (-a2 + 3a + 4) (a2 + 3a – 4)
= (-a2 + 4a – a + 4) (a2 + 4a – a – 4) [By splitting the middle term]
= [a(-a + 4) + 1(-a + 4)] [a(a + 4) – 1(a + 4)]
= [(-a + 4) (a + 1)] [(a – 1) (a + 4)]
= (4 – a) (a + 1) (a – 1) (a + 4)
27. x2 + 1/x2 – 11
Solution:
We have, x2 + 1/x2 – 11
= x2 + 1/x2 – 2 – 9
= (x2 + 1/x2 – 2 × x × 1/x) – 9
= (x – 1/x)2 – 32
= (x – 1/x + 3) (x – 1/x – 3) [As a2 – b2 = (a + b)(a – b)]
28. 4x2 + 1/4x2 + 1
Solution:
We have, 4x2 + 1/4x2 + 1
= 4x2 + 1/4x2 + 2 – 1
= [(2x)2 + (1/2x)2 + 2 × 2x × 1/2x)] – 12
= (2x + 1/2x)2 – 12
= (2x + 1/2x + 1) (2x – 1/2x – 1) [As x2 – y2 = (x + y)(x – y)]
29. 4x4 – x2 – 12x – 36
Solution:
We know, 4x4 – x2 – 12x – 36
= 4x4 – (x2 + 12x + 36)
= (2x2)2 – [x2 + 2(x)(6) + 62]
= (2x2)2 – (x + 6)2
= (2x2 + x + 6) (2x2 – x – 6) [As x2 – y2 = (x + y)(x – y)]
= (2x2 + x + 6) (2x2 – 4x + 3x – 6) [By splitting the middle term]
= (2x2 + x + 6) [2x(x – 2) + 3(x – 2)]
= (2x2 + x + 6) [(2x + 3) (x – 2)]
= (2x2 + x + 6) (2x + 3) (x – 2)
30. a2(b + c) – (b + c)3
Solution:
We have, a2(b + c) – (b + c)3
= (b + c) [a2 – (b + c)2]
= (b + c) [(a – b – c) (a + b + c)] [As x2 – y2 = (x + y)(x – y)]
= (b + c) (a – b – c) (a + b + c)
Exercise 5(D)
Factorize:
1. a3 – 27
Solution:
We have, a3 – 27
= a3 – 33
= (a – 3) [a2 + (a x 3) + 32] [As, a3 – b3 = (a – b) (a2 + ab + b2)]
= (a – 3) (a2 + 3a + 9)
2. 1 – 8a3
Solution:
We have, 1 – 8a3
= 13 – (2a)3
= (1 – 2a) [12 + (1 x 2a) + (2a)2]
= (1 – 2a) (1 + 2a + 4a2) [As, a3 – b3 = (a – b) (a2 + ab + b2)]
3. 64 – a3b3
Solution:
We have, 64 – a3b3
= 43 – (ab)3
= (4 – ab) [42 + (4 x ab) + (ab)2]
= (4 – ab) (16 + 4ab + a2b2) [As, a3 – b3 = (a – b) (a2 + ab + b2)]
4. a6 + 27b3
Solution:
We have, a6 + 27b3
= (a2)3 + (3b)3
= (a2 + 3b) [(a2)2 – (a2 x 3b) + (3b)2]
= (a2 + 3b) (a4 – 3a2b + 9b2) [As, a3 + b3 = (a + b) (a2 – ab + b2)]
5. 3x7y – 81x4y4
Solution:
We have, 3x7y – 81x4y4
= 3xy (x6 – 27x3y3)
= 3xy [(x2)3 – (3xy)3]
= 3xy (x2 – 3xy) [(x2)2 + (x2 × 3xy) + (3xy)2] [As, a3 – b3 = (a – b) (a2 + ab + b2)]
= 3xy (x2 – 3xy) (x4 + 3x3y + 9x2y2)
= 3xy. x(x – 3y). x2(x2 + 3xy + 9y2) [Taking common from terms]
= 3x4y (x – 3y) (x2 + 3xy + 9y2)
6. a3 – 27/a3
Solution:
We have, a3 – 27/a3
= a3 – (3/a)3
= (a – 3/a) [a2 + a x 3/a + (3/a)2] [As, a3 – b3 = (a – b) (a2 + ab + b2)]
= (a – 3/a) (a2 + 3 + 9/a2)
7. a3 + 0.064
Solution:
We have, a3 + 0.064
= a3 + (0.4)3
= (a + 0.4) [a2 – (a x 0.4) + 0.42]
= (a + 0.4) (a2 – 0.4a + 0.16) [As, a3 + b3 = (a + b) (a2 – ab + b2)]
8. a4 – 343a
Solution:
We have, a4 – 343a
= a (a3 – 343)
= a (a3 – 73)
= a (a – 7) [a2 + (a x 7) + 72] [As, a3 – b3 = (a – b) (a2 + ab + b2)]
= a (a – 7) (a2 + 7a + 49)
9. (x – y)3 – 8x3
Solution:
We have, (x – y)3 – 8x3
= (x – y)3 – (2x)3
= (x – y – 2x) [(x – y)2 + 2x(x – y) + (2x)2] [As, a3 – b3 = (a – b) (a2 + ab + b2)]
= (-x – y) [x2 + y2 – 2xy + 2x2 – 2xy + 4x2]
= -(x + y) [7x2 – 4xy + y2]
10. 8a3/27 – b3/8
Solution:
We have, 8a3/27 – b3/8
= (2a/3)3 – (b/2)3
= (2a/3 – b/2) [(2a/3)2 + (2a/3 x b/2) + (b/2)2] [As, a3 – b3 = (a – b) (a2 + ab + b2)]
= (2a/3 – b/2) (4a2/9 + ab/3 + b2/4)
11. a6 – b6
Solution:
We have, a6 – b6
= (a3)2 – (b3)2
= (a3 + b3) (a3 – b3) [As x2 – y2 = (x + y) (x – y)]
Now,
= [(a + b) (a2 – ab + b2)] [(a – b) (a2 + ab + b2)] [Using identities]
= (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2)
12. a6 – 7a3 – 8
Solution:
We have, a6 – 7a3 – 8
By splitting the middle term,
= a6 – 8a3 + a3 – 8
= a3(a3 – 8) + 1(a3 – 8)
= (a3 + 1) (a3 – 8)
We know that,
a3 – b3 = (a – b) (a2 + ab + b2) … (1)
a3 + b3 = (a + b) (a2 – ab + b2) … (2)
Now,
(a3 + 1) (a3 – 8)
= [(a + 1) (a2 – a + 1)] [(a – 2)(a2 + 2a + 4)] … [Using (1) and (2)]
= (a + 1) (a – 2) (a2 + 2a + 4) (a2 – a + 1)
13. a3 – 27b3 + 2a2b – 6ab2
Solution:
We have, a3 – 27b3 + 2a2b – 6ab2
= [a3 – (3b)3] + 2ab(a – 3b)
= (a – 3b) (a2 + 3ab + 9b2) + 2ab(a – 3b) [As, a3 – b3 = (a – b) (a2 + ab + b2)]
Now, taking (a – 3b) as common
= (a – 3b) [(a2 + 3ab + 9b2) + 2ab]
= (a – 3b) (a2 + 5ab + 9b2)
14. 8a3 – b3 – 4ax + 2bx
Solution:
We have, 8a3 – b3 – 4ax + 2bx
= (2a)3 – b3 – 2x(2a – b)
= (2a – b) [(2a)2 – 2ab + b2] – 2x(2a – b)
Taking (2a – b) as common,
= (2a – b) [(4a2 + 2ab + b2) – 2x]
= (2a – b) (4a2 + 2ab + b2 – 2x)
15. a – b – a3 + b3
Solution:
We have, a – b – a3 + b3
= (a – b) – (a3 – b3)
= (a – b) – [(a – b) (a2 + ab + b2)] [As, a3 – b3 = (a – b) (a2 + ab + b2)]
Now, taking (a – b) as common
= (a – b) [1 – (a2 + ab + b2)]
= (a – b) (1 – a2 – ab – b2)
16. 2x3 + 54y3 – 4x – 12y
Solution:
We have, 2x3 + 54y3 – 4x – 12y
= 2(x3 + 27y3 – 2x – 6y)
Now,
= 2 {[(x)3 + (3y)3] – 2(x + 3y)}
= 2 {[(x + 3y) (x2 – 3xy + 9y2)] – 2(x + 3y)} [As, a3 + b3 = (a + b) (a2 – ab + b2)]
= 2 (x + 3y) (x2 – 3xy + 9y2 – 2)
17. 1029 – 3x3
Solution:
We have, 1029 – 3x3
= 3(343 – x3)
= 3(73 – x3)
= 3(7 – x) (72 + 7x + x2) [As, a3 – b3 = (a – b) (a2 + ab + b2)]
= 3(7 – x) (49 + 7x + x2)
18. Show that:
(i) 133 – 53 is divisible by 8
(ii)353 + 273 is divisible by 62
Solution:
(i) We have, (133 – 53)
Now, using identity (a3 – b3) = (a – b) (a2 + ab + b2)
= (13 – 5) (132 + 13 × 5 + 52)
= 8 × (169 + 65 + 25)
Hence, the number is divisible by 8.
(ii) (353 + 273)
Now, using identity (a3 + b3) = (a + b) (a2 – ab + b2)
= (35 + 27) (352 + 35× 27 + 272)
= 62 × (352 + 35 × 27 + 272)
Hence, the number is divisible by 62.
19. Evaluate:
Solution:
Let a = 5.67 and b = 4.33
Then,
= a + b
= 5.67 + 4.33
= 10
Exercise 5(E)
Factorize:
1. x2 + 1/4x2 + 1 – 7x – 7/2x
Solution:
We have,
= [x2 + 1/(2x)2 + 2 × x × 1/(2x)] – 7 [x + 1/(2x)]
= (x + 1/2x)2 – 7(x + 1/x)
Taking out (x + 1/2x) as common,
= (x + 1/2x) (x + 1/2x – 7)
2. 9a2 + 1/9a2 – 2 – 12a + 4/3a
Solution:
3. x2 + (a2 + 1) x/a + 1
Solution:
4. x4 + y4 – 27x2y2
Solution:
We have, x4 + y4 – 27x2y2
= x4 + y4 – 2x2y2 – 25x2y2
= [(x2) + (y2) – 2x2y2] – 25x2y2
= (x2 – y2) – (5xy)2
= (x2 – y2 – 5xy) (x2 – y2 + 5xy) [As x2 – y2 = (x + y)(x – y)]
5. 4x4 + 9y4 + 11x2y2
Solution:
We have, 4x4 + 9y4 + 11x2y2
= 4x4 + 9y4 + 12x2y2 – x2y2
= (2x2)2 + (3y2)2 + 2(2x2)(3y2) – (xy)2
= (2x2 + 3y2)2 – (xy)2
= (2x2 + 3y2 – xy) (2x2 + 3y2 + xy) [As x2 – y2 = (x + y)(x – y)]
6. x2 + 1/x2 – 3
Solution:
We have, x2 + 1/x2 – 3
= x2 + 1/x2 – 2 – 1
= [x2 + 1/x2 – (2 × x × 1/x)] – 12
= (x – 1/x)2 – 12
= (x – 1/x – 1) (x – 1/x + 1) [As x2 – y2 = (x + y)(x – y)]
7. a – b – 4a2 + 4b2
Solution:
We have, a – b – 4a2 + 4b2
= (a – b) – 4(a2 – b2)
= (a – b) – 4(a – b)(a + b) [As x2 – y2 = (x + y)(x – y)]
Taking (a – b) common,
= (a – b) [1 – 4(a + b)]
= (a – b) [1 – 4a – 4b]
8. (2a – 3)2 – 2 (2a – 3) (a – 1) + (a – 1)2
Solution:
We have, (2a – 3)2 – 2(2a – 3)(a – 1) + (a – 1)2
Comparing with the identity, (a – b)2 = a2 – 2ab + b2
= [(2a – 3) – (a – 1)]2
= (2a – a – 3 + 1)2
= (a – 2)2
9. (a2 – 3a) (a2 – 3a + 7) + 10
Solution:
Let’s substitute (a2 – 3a) = x
Then the given expression becomes,
= x(x + 7) + 10
= x2 + 7x + 10
= x2 + 5x + 2x + 10 [By splitting the middle term]
= x(x + 5) + 2(x + 5)
= (x + 2) (x + 5)
Resubstituting the value of x,
= (a2 – 3a + 2) (a2 – 3a + 5)
= (a2 – 3a + 5) (a2 – 3a + 2)
= (a2 – 3a + 5) [a2 – 2a – a + 2] [By splitting the middle term]
= (a2 – 3a + 5) [a(a – 2) – 1(a – 5)]
= (a2 – 3a + 5) [(a – 1) (a – 2)]
= (a2 – 3a + 5) (a – 1) (a – 2)
10. (a2 – a) (4a2 – 4a – 5) – 6
Solution:
Let’s a2 – a = x
Then the expression becomes,
= x(4x – 5) – 6
= 4x2 – 5x – 6
= 4x2 – 8x + 3x – 6
= 4x(x – 2) + 3(x – 2)
= (4x + 3) (x – 2)
Resubstituting the value of x,
= (4a2 – 4a + 3) (a2 – a – 2)
= (4a2 – 4a + 3) (a2 – 2a + a – 2)
= (4a2 – 4a + 3) [a(a – 2) + 1(a – 2)]
= (4a2 – 4a + 3) [(a + 1) (a – 2)]
= (4a2 – 4a + 3) (a + 1) (a – 2)
11. x4 + y4 – 3x2y2
Solution:
We have, x4 + y4 – 3x2y2
= (x4 + y4 – 2x2y2) – x2y2
= (x2 – y2) – (xy)2 [As x2 – y2 = (x + y)(x – y)]
= (x2 – y2 – xy) (x2 – y2 + xy)
12. 5a2 – b2 – 4ab + 7a – 7b
Solution:
We have, 5a2 – b2 – 4ab + 7a – 7b
= 4a2 + a2 – b2 – 4ab + 7a – 7b
= a2 – b2 + 4a2 – 4ab + 7a – 7b
= (a2 – b2) + 4a(a – b) + 7(a – b)
= (a + b)(a – b) + 4a(a – b) + 7(a – b) [As x2 – y2 = (x + y)(x – y)]
= (a – b) [(a + b) + 4a + 7]
= (a – b) (5a + b + 7)
13. 12(3x – 2y)2 – 3x + 2y – 1
Solution:
We have, 12(3x – 2y)2 – 3x + 2y – 1
= 12(3x – 2y)2 – (3x – 2y) – 1
Let’s substitute (3x – 2y) = a
Then, the expression becomes
= 12a2 – a – 1
= 12a2 – 4a + 3a – 1
= 4a (3a – 1) + 1(3a – 1)
= (4a + 1) (3a – 1)
Now, resubstituting the value of ‘a’ in the above
= [4(3x – 2y) + 1] [3(3x – 2y) – 1]
= (12x – 8y + 1) (9x – 6y – 1)
14. 4(2x – 3y)2 – 8x+12y – 3
Solution:
We have, 4(2x – 3y)2 – 8x+12y – 3
= 4(2x – 3y)2 – 4(2x + 3y) – 3
Let’s substitute (2x – 3y) = a
= 4(a2) – 4a – 3
= 4a2 – 6a + 2a – 3 [By splitting the middle term]
= 2a(2a – 3) + 1(2a – 3)
= (2a – 3) (2a + 1)
Now, resubstituting the value of ‘a’ in the above
= [2(2x – 3y) – 3] [2(2x – 3y) + 1]
= (4x – 6y – 3) (4x – 6y + 1)
15. 3 – 5x + 5y – 12(x – y)2
Solution:
We have, 3 – 5x + 5y – 12(x – y)2
= 3 – 5(x – y) – 12(x – y)2
Let’s substitute (x – y) = a
= 3 – 5a – 12a2
= 3 – 9a + 4a – 12a2 [By splitting the middle term]
= 3(1 – 3a) + 4a(1 – 3a)
= (1 – 3a) (4a + 3)
Now, resubstituting the value of ‘a’ in the above
= [1 – 3(x – y)] [4(x – y) + 3]
= (1 – 3x + 3y) (4x – 4y + 3)
16. 9x 2 + 3x – 8y – 64y2
Solution:
We have, 9x 2 + 3x – 8y – 64y2
On rearranging,
= 9x 2 – 64y2 + 3x – 8y
= [(3x)2 – (8y)2] + (3x – 8y)
= (3x – 8y) (3x + 8y) + (3x – 8y) [As x2 – y2 = (x + y)(x – y)]
Taking (3x – 8y) as common,
= (3x – 8y) (3x + 8y + 1)
17. 2√3x2 + x – 5√3
Solution:
We have, 2√3x2 + x – 5√3
By splitting the middle term,
= 2√3x2 + 6x – 5x – 5√3
= 2√3x(x + √3) – 5(x + √3)
= (2√3x – 5) (x + √3)
18. ¼ (a + b)2 – 9/16 (2a – b)2
Solution:
19. 2(ab + cd) – a2 – b2 + c2 + d2
Solution:
We have, 2(ab + cd) – a2 – b2 + c2 + d2
= 2ab + 2cd – a2 – b2 + c2 + d2
On rearranging and grouping, we get
= (c2 + d2 + 2cd) – (a2 + b2 – 2ab)
= (c + d)2 – (a – b)2
= [c + d – (a – b)] [c + d + (a – b)] [As x2 – y2 = (x + y)(x – y)]
= (c + d – a + b) (c + d + a – b)
20. Find the value of:
(i) 9872 – 132
(ii) (67.8)2 – (32.2)2
(iii) [(6.7)2 – (3.3)2]/ (6.7 – 3.3)
(iv) [(18.5)2 – (6.5)2]/ (18.5 – 6.5)
Solution:
(i) We have, 9872 – 132
= (987 + 13) (987 – 13)
= 1000 x 974
= 974000
(ii) We have, (67.8)2 – (32.2)2
= (67.8 + 32.2) (67.8 – 32.2)
= 100 x 35.6
= 3560
Selina Solutions for Class 9 Maths Chapter 5- Factorisation
Chapter 5, Factorisation, contains 5 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are discussed in this chapter.
5.1 Introduction
5.2 Methods of factorisation
Type 1: Taking out the common factors
Type 2: Grouping
Type 3: Trinomial of the form ax2+bx+c (by splitting method)
Type 4: Difference of two squares
Type 5: The sum or difference of two cubes
Selina Solutions for Class 9 Maths Chapter 5- Factorisation
In Chapter 5 of Class 9, the students are taught factorisation. When a polynomial is expressed as the product of two or more expressions, each of these expressions is called a factor of the polynomial. The chapter helps the students in learning the process of writing an expression in the form of terms or brackets multiplied together, the process of factorisation. Study Chapter 5 of Selina textbook to understand more about Factorisation. Learn the Selina Solutions for Class 9 effectively to come out with excellent marks in the examinations.
