ICSE Class 9 Maths Selina Solutions Chapter 5:
Here we provide you with detailed ICSE Class 9 Maths Selina Solution Chapter 5 Factorization. Our PW Subject experts have prepared these questions according to the latest syllabus for ICSE Class 9 Maths exam.
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ICSE Class 9 Maths Selina Solutions Chapter 5
In mathematics, Factorization is writing a number or other mathematical object as the product of several factors, usually the product of smaller or simpler similar objects. For example, 3 × 5 is an integer factorization of 15 and is a polynomial factorization of x²- 4.
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ICSE Class 9 Maths Selina Solutions Chapter 5 PDF
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ICSE Class 9 Maths Selina Solutions Chapter 5 PDF
ICSE Class 9 Maths Selina Solutions Chapter 5 Factorization
Here we have provided ICSE Class 9 Maths Selina Solutions Chapter 5 Factorization for the ease of students so that they can prepare better for their ICSE Class 9 Maths Exam.
ICSE Class 9 Maths Selina Solutions Chapter 5 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(3x
2
– 2y
2
) – yz(2y
2
– 3x
2
) + zx(15x
2
– 10y
2
)
Solution:
We have, xy(3x
2
– 2y
2
) – yz(2y
2
– 3x
2
) + zx(15x
2
– 10y
2
)
Changing signs to arrive at a common term
So,
= xy(3x
2
– 2y
2
) + yz(3x
2
– 2y
2
) + zx(15x
2
– 10y
2
)
= xy(3x
2
– 2y
2
) + yz(3x
2
– 2y
2
) + 5zx(3x
2
– 2y
2
)
= (3x
2
– 2y
2
) (xy + yz + 5zx)
3. ab(a
2
+ b
2
– c
2
) – bc(c
2
– a
2
– b
2
) + ca(a
2
+ b
2
– c
2
)
Solution:
We have, ab(a
2
+ b
2
– c
2
) – bc(c
2
– a
2
– b
2
) + ca(a
2
+ b
2
– c
2
)
Changing signs to arrive at a common term
So,
= ab(a
2
+ b
2
– c
2
) + bc(a
2
+ b
2
– c
2
) + ca(a
2
+ b
2
– c
2
)
= (a
2
+ b
2
– c
2
) (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. a
3
+ a – 3a
2
– 3
Solution:
We have, a
3
+ a – 3a
2
– 3
Grouping to arrive at a common term
= a (a
2
+ 1) – 3(a
2
+ 1)
Taking common, we get
= (a
2
+ 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:
a
4
– 2a
3
– 4a + 8
Solution:
We have, a
4
– 2a
3
– 4a + 8
Grouping to arrive at a common term
= a
3
(a – 2) – 4(a – 2)
Taking common, we get
= (a
3
– 4) (a – 2)
8. ab – 2b + a
2
– 2a
Solution:
We have, ab – 2b + a
2
– 2a
Grouping to arrive at a common term
= b(a – 2) + a(a – 2)
Taking common, we get
= (b + a) (a – 2)
9. ab(x
2
+ 1) + x(a
2
+ b
2
)
Solution:
We have, ab(x
2
+ 1) + x(a
2
+ b
2
)
On expanding,
= abx
2
+ ab + a
2
x + b
2
x
Now, grouping to arrive at a common term
= abx
2
+ a
2
x + b
2
x + ab
= ax(bx + a) + b(bx + a)
Taking common, we get
= (ax + b) (bx + a)
10. a
2
+ b – ab – a
Solution:
We have, a
2
+ b – ab – a
Grouping to arrive at a common term
= a
2
– 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,
= a
2
x
2
+ b
2
y
2
+ 2abxy + b
2
x
2
+ a
2
y
2
– 2abxy
= a
2
x
2
+ b
2
y
2
+ b
2
x
2
+ a
2
y
2
Rearranging terms, we get
= a
2
x
2
+ b
2
x
2
+ a
2
y
2
+ b
2
y
2
Taking common, we get
= x
2
(a
2
+ b
2
) + y
2
(a
2
+ b
2
)
= (x
2
+ y
2
) (a
2
+ b
2
)
12. a
2
x
2
+ (ax
2
+ 1)x + a
Solution:
We have, a
2
x
2
+ (ax
2
+ 1)x + a
Regrouping the terms, we have
= a
2
x
2
+ a + (ax
2
+ 1)x
= a(ax
2
+ 1) + x(ax
2
+ 1)
Taking common, we get
= (ax
2
+ 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. y
2
– (a + b) y + ab
Solution:
We have, y
2
– (a + b) y + ab
On expanding,
= y
2
– ay – by + ab
= (y
2
– ay) – by + ab
Taking ‘y’ and ‘b’ common from the group, we get
= y(y – a) – b(y – a)
= (y – a) (y – b)
16. a
2
+ 1/a
2
– 2 – 3a + 3/a
Solution:
We have, a
2
+ 1/a
2
– 2 – 3a + 3/a
On grouping terms, we get
= (a
2
– 2 + 1/a
2
) – 3a + 3/a
= [a
2
– (2 x a x 1/a) + 1/a
2
] – 3(a – 1/a)
= (a – 1/a)
2
– 3(a – 1/a) {Since, (x – y)
2
= x
2
– 2xy + y
2
}
Taking (a – 1/a) as common, we get
= (a – 1/a) [(a – 1/a) – 3]
= (a – 1/a) (a – 1/a – 3)
17. x
2
+ y
2
+ x + y + 2xy
Solution:
We have, x
2
+ y
2
+ x + y + 2xy
On rearranging terms, we get
= (x
2
+ y
2
+ 2xy) + (x + y) {Since, (x + y)
2
= x
2
+ 2xy + y
2
}
Now,
= (x + y)
2
+ (x + y)
= (x + y)(x + y + 1)
18. a
2
+ 4b
2
– 3a + 6b – 4ab
Solution:
We have, a
2
+ 4b
2
– 3a + 6b – 4ab
On rearranging terms, we get
= a
2
+ 4b
2
– 4ab – 3a + 6b
Now,
= a
2
+ (2b)
2
– 2 × a × (2b) – 3(a – 2b) {Since, (a – b)
2
= a
2
– 2ab + b
2
}
= (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. a
2
+ 10a + 24
Solution:
We have, a
2
+ 10a + 24
By splitting the middle term, we get
= a
2
+ 6a + 4a + 24
= a(a + 6) + 4(a + 6)
= (a + 4) (a + 6)
2.
a
2
– 3a – 40
Solution:
We have, a
2
– 3a – 40
By splitting the middle term, we get
= a
2
– 8a + 5a – 40
= a(a – 8) + 5(a – 8)
= (a + 5) (a – 8)
3. 1 – 2a – 3a
2
Solution:
We have, 1 – 2a – 3a
2
By splitting the middle term, we get
= 1 – 3a + a – 3a
2
= 1(1 – 3a) + a(1 – 3a)
= (1 + a) (1 – 3a)
4. x
2
– 3ax – 88a
2
Solution:
We have, x
2
– 3ax – 88a
2
By splitting the middle term, we get
= x
2
– 11ax + 8ax – 88a
2
= x(x – 11a) + 8a(x – 11a)
= (x + 8a) (x – 11a)
5. 6a
2
– a – 15
Solution:
We have, 6a
2
– a – 15
By splitting the middle term, we get
= 6a
2
+ 9a – 10a – 15
= 3a(2a + 3) – 5(2a + 3)
= (3a – 5) (2a + 3)
6. 24a
3
+ 37a
2
– 5a
Solution:
We have, 24a
3
+ 37a
2
– 5a
Taking ‘a’ common from all
= a (24a
2
+ 37a – 5)
= a (24a
2
+ 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,
= 3a
2
– 2a – 1
By splitting the middle term, we get
= 3a
2
– 3a + a – 1
= 3a(a – 1) + 1(a – 1)
= (3a + 1) (a – 1)
8. a
2
b
2
+ 8ab – 9
Solution:
We have, a
2
b
2
+ 8ab – 9
By splitting the middle term, we get
= a
2
b
2
+ 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 – 7a
2
By splitting the middle term, we get
= 3 + 3a – 7a – 7a
2
= 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
= x
2
– 3x – 4
By splitting the middle term, we get
= x
2
– 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 – 3x
2
By splitting the middle term, we get
= 1 – 3x + x – 3x
2
= 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. 3a
2
– 1 – 2a
Solution:
We have, 3a
2
– 1 – 2a
Rearranging,
= 3a
2
– 2a – 1
By splitting the middle term, we get
= 3a
2
– 3a + a – 1
= 3a(a – 1) + 1(a – 1)
= (3a + 1) (a – 1)
13. x
2
+ 3x + 2 + ax + 2a
Solution:
We have, x
2
+ 3x + 2 + ax + 2a
By splitting the middle term, we get
= (x
2
+ 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
= a
2
+ 3a – 10
By splitting the middle term, we get
= a
2
+ 5a – 2a – 10
= a(a + 5) – 2(a + 5)
= (a – 2) (a + 5)
= (3x – 2y + 5) (3x – 2y – 2)
15. 5 – (3a
2
– 2a) (6 – 3a
2
+ 2a)
Solution:
Given, 5 – (3a
2
– 2a) (6 – 3a
2
+ 2a)
= 5 – (3a
2
– 2a) [6 – (3a
2
– 2a)]
Let’s substitute (3a
2
– 2a) = x
And, the expression becomes,
= 5 – x(6 – x)
= 5 – 6x – x
2
= 5 – 5x – x – x
2
= 5(1 – x) – x(1 – x)
= (1 – x) (5 – x)
= (x – 1) (x – 5)
= (3a
2
– 2a – 1) (3a
2
– 2a – 5)
Now,
= (3a
2
– 3a + a – 1) (3a
2
+ 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 + a
2
Solution:
We have, 1/35 + 12a/35 + a
2
Taking common,
= 1/35 (1 + 12a + 35a
2
)
= 1/35 (35a
2
+ 12a + 1)
= 1/35 (35a
2
+ 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. (x
2
– 3x) (x
2
– 3x – 1) – 20.
Solution:
We have, (x
2
– 3x) (x
2
– 3x – 1) – 20
= (x
2
– 3x)[(x
2
– 3x) – 1] – 20
Let’s
= a[a – 1] – 20 ….(Taking x
2
– 3x = a)
= a
2
– a – 20
= a
2
– 5a + 4a – 20
= a(a – 5) + 4(a – 5)
= (a – 5)(a + 4)
= (x
2
– 3x – 5)(x
2
– 3x + 4)
18. Find each trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
(i) x
2
– 3x – 54
(ii) 2x
2
– 7x – 15
(iii) 2x
2
+ 2x – 75
(iv) 3x
2
+ 4x – 10
(v) x(2x – 1) – 1
Solution:
(i) Given, x
2
– 3x – 54
On comparing with the general form ax
2
+ bx + c, we get
a = 1, b = -3 and c = -54
So, b
2
– 4ac = (-3)
2
– 4(1)(-54) = 9 + 216 = 225
225 is a perfect square
Thus, x
2
– 3x – 54 is factorisable
Now,
x
2
– 3x – 54 = x
2
– 9x + 6x – 54
= x(x – 9) + 6(x – 9)
= (x + 6) (x – 9)
(ii) Given, 2x
2
– 7x – 15
On comparing with the general form ax
2
+ bx + c, we get
a = 2, b = -7 and c = -15
So, b
2
– 4ac = (-7)
2
– 4(2)(-15) = 49 + 120 = 169
169 is a perfect square
Thus, 2x
2
– 7x – 15 is factorisable
Now,
2x
2
– 7x – 15 = 2x
2
– 10x + 3x – 15
= 2x(x – 5) + 3(x – 5)
= (2x + 3) (x – 5)
(iii) Given, 2x
2
+ 2x – 75
On comparing with the general form ax
2
+ bx + c, we get
a = 2, b = 2 and c = -75
So, b
2
– 4ac = (2)
2
– 4(2)(-75) = 4 + 600 = 604
604 is not a perfect square
Thus, 2x
2
+ 2x – 75 is not factorizable
(iv) Given, 3x
2
+ 4x – 10
On comparing with the general form ax
2
+ bx + c, we get
a = 3, b = 4 and c = -10
So, b
2
– 4ac = (4)
2
– 4(3)(-10) = 16 + 120 = 136
136 is a not perfect square
Thus, 3x
2
+ 4x – 10 is not factorizable
(v) Given, x(2x – 1) – 1
= 2x
2
– x – 1
On comparing with the general form ax
2
+ bx + c, we get
a = 2, b = -1 and c = -1
So, b
2
– 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 = 2x
2
– x – 1
= 2x
2
– 2x + x – 1
= 2x(x – 1) + 1(x – 1)
= (2x + 1) (x – 1)
19. Factorise:
(i) 4√3x
2
+ 5x – 2√3
(ii) 7√2x
2
– 10x – 4√2
Solution:
(i) We have, 4√3x
2
+ 5x – 2√3
By splitting the middle term, we get
= 4√3x
2
+ 8x – 3x – 2√3
= 4x(√3x + 2) – √3(√3x + 2)
= (4x – √3) (√3x + 2)
(ii) We have, 7√2x
2
– 10x – 4√2
By splitting the middle term, we get
= 7√2x
2
– 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 12x
2
– 35x + 25.
Solution:
We have, 12x
2
– 35x + 25
By splitting the middle term, we get
= 12x
2
– 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. 25a
2
– 9b
2
Solution:
We have, 25a
2
– 9b
2
= (5a)
2
– (3b)
2
= (5a + 3b) (5a – 3b) [As x
2
– y
2
= (x + y)(x – y)
2. a
2
– (2a + 3b)
2
Solution:
We have, a
2
– (2a + 3b)
2
= [a – (2a + 3b)] [a + (2a + 3b)] [As x
2
– y
2
= (x + y)(x – y)]
= (a – 2a – 3b) (a + 2a + 3b)
= (-a – 3b) (3a + 3b)
= -3(a + 3b) (a + b)
3. a
2
– 81(b-c)
2
Solution:
We have, a
2
– 81(b-c)
2
= a
2
– [9(b – c)]
2
= [a – 9(b – c)] [a + 9(b – c)]] [As x
2
– y
2
= (x + y)(x – y)]
= (a – 9b + 9c) (a + 9b – 9c)
4. 25(2a – b)
2
– 81b
2
Solution:
We have, 25(2a – b)
2
– 81b
2
= [5(2a – b)]
2
– (9b)
2
= [5(2a – b) – 9b] [5(2a – b) + 9b] [As x
2
– y
2
= (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. 50a
3
– 2a
Solution:
We have, 50a
3
– 2a
= 2a(25a
2
– 1)
= 2a[(5a)
2
– 1
2
]
= 2a(5a – 1)(5a + 1) [As x
2
– y
2
= (x + y)(x – y)
6. 4a
2
b – 9b
3
Solution:
We have, 4a
2
b – 9b
3
= b(4a
2
– 9b
2
)
= b[(2a)
2
– (3b)
2
]
= b[(2a + 3b) (2a – 3b)] [As x
2
– y
2
= (x + y)(x – y)]
= b(2a + 3b)(2a – 3b)
7. 3a
5
– 108a
3
Solution:
We have, 3a
5
– 108a
3
= 3a
3
(a
2
– 36)
= 3a
3
(a
2
– 6
2
)
= 3a
3
(a – 6) (a + 6)
[As x
2
– y
2
= (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 x
2
– y
2
= (x + y)(x – y)]
= [3a – 6 – 4a – 8] [3a – 6 + 4a + 8]
= [-a – 14] [7a + 2]
= -(a + 14) (7a + 2)
9. a
4
– 1
Solution:
We have, a
4
– 1
= (a
2
)
2
– 1
2
= (a
2
– 1) (a
2
+ 1) [As x
2
– y
2
= (x + y)(x – y)]
= [(a – 1)(a + 1)] (a
2
+ 1)
= (a – 1)(a + 1)(a
2
+ 1)
10. a
3
+ 2a
2
– a – 2
Solution:
We have, a
3
+ 2a
2
– a – 2
= a
2
(a + 2) – 1(a + 2)
= (a
2
– 1) (a + 2)
= (a – 1) (a + 1) (a + 2) [As x
2
– y
2
= (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 x
2
– y
2
= (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)
= a
2
– a – b
2
+ b
= (a
2
– b
2
) – (a – b)
= (a + b) (a – b) – (a – b) [As x
2
– y
2
= (x + y)(x – y)]
= (a – b) [(a + b) – 1]
= (a – b) (a + b – 1)
13. 4a
2
– (4b
2
+ 4bc + c
2
)
Solution:
We know, 4a
2
– (4b
2
+ 4bc + c
2
)
= (2a)
2
– [(2b)
2
+ 2(2b)(c) + c
2
]
= (2a)
2
– (2b + c)
2
= (2a – 2b – c) (2a + 2b + c) [As x
2
– y
2
= (x + y)(x – y)]
14. 4a
2
– 49b
2
+ 2a – 7b
Solution:
We know, 4a
2
– 49b
2
+ 2a – 7b
= (2a)
2
– (7b)
2
+ (2a – 7b)
= [(2a – 7b) (2a + 7b)] + (2a – 7b) [As x
2
– y
2
= (x + y)(x – y)]
= (2a – 7b) [(2a + 7b) + 1]
= (2a – 7b) (2a + 7b + 1)
15. 9a
2
+ 3a – 8b – 64b
2
Solution:
We have, 9a
2
+ 3a – 8b – 64b
2
= 9a
2
– 64b
2
+ 3a – 8b
= (3a)
2
– (8b)
2
+ (3a – 8b)
= [(3a – 8b) (3a + 8b)] + (3a – 8b) [As x
2
– y
2
= (x + y)(x – y)]
= (3a – 8b) [(3a + 8b) + 1]
= (3a – 8b) (3a + 8b + 1)
16. 4a
2
– 12a + 9 – 49b
2
Solution:
We have, 4a
2
– 12a + 9 – 49b
2
= [(2a)
2
– 2(2a)(3) + 3
2
] – (7b)
2
= (2a – 3)
2
– (7b)
2
= (2a – 7b – 3) (2a + 7b – 3) [As x
2
– y
2
= (x + y)(x – y)]
17. 4xy – x
2
– 4y
2
+ z
2
Solution:
We have, 4xy – x
2
– 4y
2
+ z
2
On rearranging,
= z
2
– x
2
– 4y
2
+ 4xy
= z
2
– (x
2
+ 4y
2
– 4xy)
= z
2
– (x – 2y)
2
= (z – x + 2y) (z + x – 2y) [As x
2
– y
2
= (x + y)(x – y)]
18. a
2
+ b
2
– c
2
– d
2
+ 2ab – 2cd
Solution:
We have, a
2
+ b
2
– c
2
– d
2
+ 2ab – 2cd
On rearranging,
= a
2
+ 2ab + b
2
– c
2
– d
2
– 2cd
= (a
2
+ 2ab + b
2
) – (c
2
+ d
2
+ 2cd)
= (a + b)
2
– (c + d)
2
= (a + b + c + d) (a + b – c – d) [As x
2
– y
2
= (x + y)(x – y)]
19. 4x
2
– 12ax – y
2
– z
2
– 2yz + 9a
2
Solution:
We have, 4x
2
– 12ax – y
2
– z
2
– 2yz + 9a
2
On rearranging,
= 4x
2
– 12ax + 9a
2
– y
2
– z
2
– 2yz
= (4x
2
– 12ax + 9a
2
) – (y
2
+ z
2
+ 2yz)
= (2x – 3a)
2
– (y + z)
2
= (2x – 3a + y + z) (2x – 3a – y – z) [As x
2
– y
2
= (x + y)(x – y)]
20. (a
2
– 1) (b
2
– 1) + 4ab
Solution:
We have, (a
2
– 1) (b
2
– 1) + 4ab
By cross multiplying and expanding, we get
= (1 – a
2
– b
2
+ a
2
b
2
) + 4ab
On manipulating,
= (a
2
b
2
+ 1 + 2ab) – (a
2
+ b
2
– 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 x
2
– y
2
= (x + y)(x – y)]
21. x
4
+ x
2
+ 1
Solution:
We have, x
4
+ x
2
+ 1
= x
4
+ 2x
2
+ 1 – x
2
= (x
2
+ 1)
2
– x
2
[As a
2
– b
2
= (a + b)(a – b)]
= (x
2
+ 1 – x) (x
2
+ 1 + x)
22. (a
2
+ b
2
– 4c
2
)
2
– 4a
2
b
2
Solution:
We have, (a
2
+ b
2
– 4c
2
)
2
– 4a
2
b
2
= (a
2
+ b
2
– 4c
2
)
2
– (2ab)
2
= [(a
2
+ b
2
– 4c
2
) + (2ab)] [(a
2
+ b
2
– 4c
2
) – (2ab)] [As x
2
– y
2
= (x + y)(x – y)]
= [(a
2
+ b
2
+ 2ab) – 4c
2
] [(a
2
+ b
2
– 2ab) – 4c
2
]
= [(a + b)
2
– (2c)
2
] [(a – b)
2
– (2c)
2
]
= [(a + b – 2c) (a + b + 2c)] [(a – b – 2c) (a – b + 2c)] [As x
2
– y
2
= (x + y)(x – y)]
= (a + b – 2c) (a + b + 2c) (a – b – 2c) (a – b + 2c)
23. (x
2
+ 4y
2
– 9z
2
)
2
– 16x
2
y
2
Solution:
We have, (x
2
+ 4y
2
– 9z
2
)
2
– 16x
2
y
2
= (x
2
+ 4y
2
– 9z
2
)
2
– (4xy)
2
= (x
2
+ 4y
2
– 9z
2
– 4xy) (x
2
+ 4y
2
– 9z
2
+ 4xy) [As x
2
– y
2
= (x + y)(x – y)]
= [(x
2
– 4xy + 4y
2
) – 9z
2
] [(x
2
+ 4xy + 4y
2
) – 9z
2
]
= [(x – 2y)
2
– (3z)
2
] [(x + 2y)
2
– (3z)
2
]
= [(x – 2y + 3z) (x – 2y – 3z)] [(x +2y + 3z) (x + 2y – 3z)] [As x
2
– y
2
= (x + y)(x – y)]
= (x – 2y + 3z) (x – 2y – 3z)] [(x +2y + 3z) (x + 2y – 3z)
24. (a + b)
2
– a
2
+ b
2
Solution:
We have, (a + b)
2
– a
2
+ b
2
On expanding,
= (a
2
+ 2ab + b
2
) – a
2
+ b
2
= 2b
2
+ 2ab
= 2b (b + a)
25. a
2
– b
2
– (a + b)
2
Solution:
We have, a
2
– b
2
– (a + b)
On expanding,
= a
2
– b
2
– (a
2
+ b
2
+ 2ab)
= a
2
– b
2
– a
2
– b
2
– 2ab
= -2b
2
– 2ab
= -2b(b + a)
26. 9a
2
– (a
2
– 4)
2
Solution:
We have, 9a
2
– (a
2
– 4)
2
= (3a)
2
– (a
2
– 4)
2
= [3a – (a
2
– 4)] [3a + (a
2
– 4)] [As x
2
– y
2
= (x + y)(x – y)]
= [3a – (a
2
– 2
2
)] [3a + (a
2
– 2
2
)]
= (3a – a
2
+ 4) (3a + a
2
– 4)
= (-a
2
+ 3a + 4) (a
2
+ 3a – 4)
= (-a
2
+ 4a – a + 4) (a
2
+ 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. x
2
+ 1/x
2
– 11
Solution:
We have, x
2
+ 1/x
2
– 11
= x
2
+ 1/x
2
– 2 – 9
= (x
2
+ 1/x
2
– 2 × x × 1/x) – 9
= (x – 1/x)
2
– 3
2
= (x – 1/x + 3) (x – 1/x – 3) [As a
2
– b
2
= (a + b)(a – b)]
28. 4x
2
+ 1/4x
2
+ 1
Solution:
We have, 4x
2
+ 1/4x
2
+ 1
= 4x
2
+ 1/4x
2
+ 2 – 1
= [(2x)
2
+ (1/2x)
2
+ 2 × 2x × 1/2x)] – 1
2
= (2x + 1/2x)
2
– 1
2
= (2x + 1/2x + 1) (2x – 1/2x – 1) [As x
2
– y
2
= (x + y)(x – y)]
29. 4x
4
– x
2
– 12x – 36
Solution:
We know, 4x
4
– x
2
– 12x – 36
= 4x
4
– (x
2
+ 12x + 36)
= (2x
2
)
2
– [x
2
+ 2(x)(6) + 6
2
]
= (2x
2
)
2
– (x + 6)
2
= (2x
2
+ x + 6) (2x
2
– x – 6) [As x
2
– y
2
= (x + y)(x – y)]
= (2x
2
+ x + 6) (2x
2
– 4x + 3x – 6) [By splitting the middle term]
= (2x
2
+ x + 6) [2x(x – 2) + 3(x – 2)]
= (2x
2
+ x + 6) [(2x + 3) (x – 2)]
= (2x
2
+ x + 6) (2x + 3) (x – 2)
30. a
2
(b + c) – (b + c)
3
Solution:
We have, a
2
(b + c) – (b + c)
3
= (b + c) [a
2
– (b + c)
2
]
= (b + c) [(a – b – c) (a + b + c)] [As x
2
– y
2
= (x + y)(x – y)]
= (b + c) (a – b – c) (a + b + c)
Exercise 5(D)
Factorize:
1. a
3
– 27
Solution:
We have, a
3
– 27
= a
3
– 3
3
= (a – 3) [a
2
+ (a x 3) + 3
2
] [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
= (a – 3) (a
2
+ 3a + 9)
2. 1 – 8a
3
Solution:
We have, 1 – 8a
3
= 1
3
– (2a)
3
= (1 – 2a) [1
2
+ (1 x 2a) + (2a)
2
]
= (1 – 2a) (1 + 2a + 4a
2
) [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
3. 64 – a
3
b
3
Solution:
We have, 64 – a
3
b
3
= 4
3
– (ab)
3
= (4 – ab) [4
2
+ (4 x ab) + (ab)
2
]
= (4 – ab) (16 + 4ab + a
2
b
2
) [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
4. a
6
+ 27b
3
Solution:
We have, a
6
+ 27b
3
= (a
2
)
3
+ (3b)
3
= (a
2
+ 3b) [(a
2
)
2
– (a
2
x 3b) + (3b)
2
]
= (a
2
+ 3b) (a
4
– 3a
2
b + 9b
2
) [As, a
3
+ b
3
= (a + b) (a
2
– ab + b
2
)]
5. 3x
7
y – 81x
4
y
4
Solution:
We have, 3x
7
y – 81x
4
y
4
= 3xy (x
6
– 27x
3
y
3
)
= 3xy [(x
2
)
3
– (3xy)
3
]
= 3xy (x
2
– 3xy) [(x
2
)
2
+ (x
2
× 3xy) + (3xy)
2
] [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
= 3xy (x
2
– 3xy) (x
4
+ 3x
3
y + 9x
2
y
2
)
= 3xy. x(x – 3y). x
2
(x
2
+ 3xy + 9y
2
) [Taking common from terms]
= 3x
4
y (x – 3y) (x
2
+ 3xy + 9y
2
)
6. a
3
– 27/a
3
Solution:
We have, a
3
– 27/a
3
= a
3
– (3/a)
3
= (a – 3/a) [a
2
+ a x 3/a + (3/a)
2
] [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
= (a – 3/a) (a
2
+ 3 + 9/a
2
)
7. a
3
+ 0.064
Solution:
We have, a
3
+ 0.064
= a
3
+ (0.4)
3
= (a + 0.4) [a
2
– (a x 0.4) + 0.4
2
]
= (a + 0.4) (a
2
– 0.4a + 0.16) [As, a
3
+ b
3
= (a + b) (a
2
– ab + b
2
)]
8. a
4
– 343a
Solution:
We have, a
4
– 343a
= a (a
3
– 343)
= a (a
3
– 7
3
)
= a (a – 7) [a
2
+ (a x 7) + 7
2
] [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
= a (a – 7) (a
2
+ 7a + 49)
9. (x – y)
3
– 8x
3
Solution:
We have, (x – y)
3
– 8x
3
= (x – y)
3
– (2x)
3
= (x – y – 2x) [(x – y)
2
+ 2x(x – y) + (2x)
2
] [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
= (-x – y) [x
2
+ y
2
– 2xy + 2x
2
– 2xy + 4x
2
]
= -(x + y) [7x
2
– 4xy + y
2
]
10. 8a
3
/27 – b
3
/8
Solution:
We have, 8a
3
/27 – b
3
/8
= (2a/3)
3
– (b/2)
3
= (2a/3 – b/2) [(2a/3)
2
+ (2a/3 x b/2) + (b/2)
2
] [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
= (2a/3 – b/2) (4a
2
/9 + ab/3 + b
2
/4)
11. a
6
– b
6
Solution:
We have, a
6
– b
6
= (a
3
)
2
– (b
3
)
2
= (a
3
+ b
3
) (a
3
– b
3
) [As x
2
– y
2
= (x + y) (x – y)]
Now,
= [(a + b) (a
2
– ab + b
2
)] [(a – b) (a
2
+ ab + b
2
)] [Using identities]
= (a + b) (a – b) (a
2
– ab + b
2
) (a
2
+ ab + b
2
)
12. a
6
– 7a
3
– 8
Solution:
We have, a
6
– 7a
3
– 8
By splitting the middle term,
= a
6
– 8a
3
+ a
3
– 8
= a
3
(a
3
– 8) + 1(a
3
– 8)
= (a
3
+ 1) (a
3
– 8)
We know that,
a
3
– b
3
= (a – b) (a
2
+ ab + b
2
) … (1)
a
3
+ b
3
= (a + b) (a
2
– ab + b
2
) … (2)
Now,
(a
3
+ 1) (a
3
– 8)
= [(a + 1) (a
2
– a + 1)] [(a – 2)(a
2
+ 2a + 4)] … [Using (1) and (2)]
= (a + 1) (a – 2) (a
2
+ 2a + 4) (a
2
– a + 1)
13. a
3
– 27b
3
+ 2a
2
b – 6ab
2
Solution:
We have, a
3
– 27b
3
+ 2a
2
b – 6ab
2
= [a
3
– (3b)
3
] + 2ab(a – 3b)
= (a – 3b) (a
2
+ 3ab + 9b
2
) + 2ab(a – 3b) [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
Now, taking (a – 3b) as common
= (a – 3b) [(a
2
+ 3ab + 9b
2
) + 2ab]
= (a – 3b) (a
2
+ 5ab + 9b
2
)
14. 8a
3
– b
3
– 4ax + 2bx
Solution:
We have, 8a
3
– b
3
– 4ax + 2bx
= (2a)
3
– b
3
– 2x(2a – b)
= (2a – b) [(2a)
2
– 2ab + b
2
] – 2x(2a – b)
Taking (2a – b) as common,
= (2a – b) [(4a
2
+ 2ab + b
2
) – 2x]
= (2a – b) (4a
2
+ 2ab + b
2
– 2x)
15. a – b – a
3
+ b
3
Solution:
We have, a – b – a
3
+ b
3
= (a – b) – (a
3
– b
3
)
= (a – b) – [(a – b) (a
2
+ ab + b
2
)] [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
Now, taking (a – b) as common
= (a – b) [1 – (a
2
+ ab + b
2
)]
= (a – b) (1 – a
2
– ab – b
2
)
16. 2x
3
+ 54y
3
– 4x – 12y
Solution:
We have, 2x
3
+ 54y
3
– 4x – 12y
= 2(x
3
+ 27y
3
– 2x – 6y)
Now,
= 2 {[(x)
3
+ (3y)
3
] – 2(x + 3y)}
= 2 {[(x + 3y) (x
2
– 3xy + 9y
2
)] – 2(x + 3y)} [As, a
3
+ b
3
= (a + b) (a
2
– ab + b
2
)]
= 2 (x + 3y) (x
2
– 3xy + 9y
2
– 2)
17. 1029 – 3x
3
Solution:
We have, 1029 – 3x
3
= 3(343 – x
3
)
= 3(7
3
– x
3
)
= 3(7 – x) (7
2
+ 7x + x
2
) [As, a
3
– b
3
= (a – b) (a
2
+ ab + b
2
)]
= 3(7 – x) (49 + 7x + x
2
)
18. Show that:
(i) 13
3
– 5
3
is divisible by 8
(ii)35
3
+ 27
3
is divisible by 62
Solution:
(i) We have, (13
3
– 5
3
)
Now, using identity (a
3
– b
3
) = (a – b) (a
2
+ ab + b
2
)
= (13 – 5) (13
2
+ 13 × 5 + 5
2
)
= 8 × (169 + 65 + 25)
Hence, the number is divisible by 8.
(ii) (35
3
+ 27
3
)
Now, using identity (a
3
+ b
3
) = (a + b) (a
2
– ab + b
2
)
= (35 + 27) (35
2
+ 35× 27 + 27
2
)
= 62 × (35
2
+ 35 × 27 + 27
2
)
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. x
2
+ 1/4x
2
+ 1 – 7x – 7/2x
Solution:
We have,
= [x
2
+ 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. 9a
2
+ 1/9a
2
– 2 – 12a + 4/3a
Solution:
3. x
2
+ (a
2
+ 1) x/a + 1
Solution:
4. x
4
+ y
4
– 27x
2
y
2
Solution:
We have, x
4
+ y
4
– 27x
2
y
2
= x
4
+ y
4
– 2x
2
y
2
– 25x
2
y
2
= [(x
2
) + (y
2
) – 2x
2
y
2
] – 25x
2
y
2
= (x
2
– y
2
) – (5xy)
2
= (x
2
– y
2
– 5xy) (x
2
– y
2
+ 5xy) [As x
2
– y
2
= (x + y)(x – y)]
5. 4x
4
+ 9y
4
+ 11x
2
y
2
Solution:
We have, 4x
4
+ 9y
4
+ 11x
2
y
2
= 4x
4
+ 9y
4
+ 12x
2
y
2
– x
2
y
2
= (2x
2
)
2
+ (3y
2
)
2
+ 2(2x
2
)(3y
2
) – (xy)
2
= (2x
2
+ 3y
2
)
2
– (xy)
2
= (2x
2
+ 3y
2
– xy) (2x
2
+ 3y
2
+ xy) [As x
2
– y
2
= (x + y)(x – y)]
6. x
2
+ 1/x
2
– 3
Solution:
We have, x
2
+ 1/x
2
– 3
= x
2
+ 1/x
2
– 2 – 1
= [x
2
+ 1/x
2
– (2 × x × 1/x)] – 1
2
= (x – 1/x)
2
– 1
2
= (x – 1/x – 1) (x – 1/x + 1) [As x
2
– y
2
= (x + y)(x – y)]
7. a – b – 4a
2
+ 4b
2
Solution:
We have, a – b – 4a
2
+ 4b
2
= (a – b) – 4(a
2
– b
2
)
= (a – b) – 4(a – b)(a + b) [As x
2
– y
2
= (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
= a
2
– 2ab + b
2
= [(2a – 3)
– (a – 1)]
2
= (2a – a – 3 + 1)
2
= (a – 2)
2
9. (a
2
– 3a) (a
2
– 3a + 7) + 10
Solution:
Let’s substitute (a
2
– 3a) = x
Then the given expression becomes,
= x(x + 7) + 10
= x
2
+ 7x + 10
= x
2
+ 5x + 2x + 10 [By splitting the middle term]
= x(x + 5) + 2(x + 5)
= (x + 2) (x + 5)
Resubstituting the value of x,
= (a
2
– 3a + 2) (a
2
– 3a + 5)
= (a
2
– 3a + 5) (a
2
– 3a + 2)
= (a
2
– 3a + 5) [a
2
– 2a – a + 2] [By splitting the middle term]
= (a
2
– 3a + 5) [a(a – 2) – 1(a – 5)]
= (a
2
– 3a + 5) [(a – 1) (a – 2)]
= (a
2
– 3a + 5) (a – 1) (a – 2)
10. (a
2
– a) (4a
2
– 4a – 5) – 6
Solution:
Let’s a
2
– a = x
Then the expression becomes,
= x(4x – 5) – 6
= 4x
2
– 5x – 6
= 4x
2
– 8x + 3x – 6
= 4x(x – 2) + 3(x – 2)
= (4x + 3) (x – 2)
Resubstituting the value of x,
= (4a
2
– 4a + 3) (a
2
– a – 2)
= (4a
2
– 4a + 3) (a
2
– 2a + a – 2)
= (4a
2
– 4a + 3) [a(a – 2) + 1(a – 2)]
= (4a
2
– 4a + 3) [(a + 1) (a – 2)]
= (4a
2
– 4a + 3) (a + 1) (a – 2)
11. x
4
+ y
4
– 3x
2
y
2
Solution:
We have, x
4
+ y
4
– 3x
2
y
2
= (x
4
+ y
4
– 2x
2
y
2
) – x
2
y
2
= (x
2
– y
2
) – (xy)
2
[As x
2
– y
2
= (x + y)(x – y)]
= (x
2
– y
2
– xy) (x
2
– y
2
+ xy)
12. 5a
2
– b
2
– 4ab + 7a – 7b
Solution:
We have, 5a
2
– b
2
– 4ab + 7a – 7b
= 4a
2
+ a
2
– b
2
– 4ab + 7a – 7b
= a
2
– b
2
+ 4a
2
– 4ab + 7a – 7b
= (a
2
– b
2
) + 4a(a – b) + 7(a – b)
= (a + b)(a – b) + 4a(a – b) + 7(a – b) [As x
2
– y
2
= (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
= 12a
2
– a – 1
= 12a
2
– 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(a
2
) – 4a – 3
= 4a
2
– 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 – 12a
2
= 3 – 9a + 4a – 12a
2
[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 – 64y
2
Solution:
We have, 9x
2
+ 3x – 8y – 64y
2
On rearranging,
= 9x
2
– 64y
2
+ 3x – 8y
= [(3x)
2
– (8y)
2
] + (3x – 8y)
= (3x – 8y) (3x + 8y) + (3x – 8y) [As x
2
– y
2
= (x + y)(x – y)]
Taking (3x – 8y) as common,
= (3x – 8y) (3x + 8y + 1)
17. 2√3x
2
+ x – 5√3
Solution:
We have, 2√3x
2
+ x – 5√3
By splitting the middle term,
= 2√3x
2
+ 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) – a
2
– b
2
+ c
2
+ d
2
Solution:
We have, 2(ab + cd) – a
2
– b
2
+ c
2
+ d
2
= 2ab + 2cd – a
2
– b
2
+ c
2
+ d
2
On rearranging and grouping, we get
= (c
2
+ d
2
+ 2cd) – (a
2
+ b
2
– 2ab)
= (c + d)
2
– (a – b)
2
= [c + d – (a – b)] [c + d + (a – b)] [As x
2
– y
2
= (x + y)(x – y)]
= (c + d – a + b) (c + d + a – b)
20. Find the value of:
(i) 987
2
– 13
2
(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, 987
2
– 13
2
= (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
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