## RD Sharma Solutions for Class 8 Chapter 7 Factorization Free Online

EXERCISE 7.1 PAGE NO: 7.3
Find the greatest common factor (GCF/HCF) of the following polynomials: (1-14)
1. 2x2 and 12x2
Solution:
We know that the numerical coefficients of given numerical are 2 and 12
The greatest common factor of 2 and 12 is 2
The common literals appearing in given monomial is x
The smallest power of x in two monomials is 2
The monomial of common literals with smallest power is x2
∴ The greatest common factor = 2x2
2. 6x3y and 18x2y3
Solution:
We know that the numerical coefficients of given numerical are 6 and18
The greatest common factor of 6 and 18 is 6
Common literals appearing in given numerical are x and y
Smallest power of x in three monomial is 2
Smallest power of y in three monomial is 1
Monomial of common literals with smallest power is x2y
∴ The greatest common factor = 6x2y
3. 7x, 21x2 and 14xy2
Solution:
We know that the numerical coefficients of given numerical are 7, 21 and 14
Greatest common factor of 7, 21 and 14 is 7
Common literals appearing in given numerical are x and y
Smallest power of x in three monomials is 1
Smallest power of y in three monomials is 0
Monomials of common literals with smallest power is x
∴ The greatest common factor = 7x
4. 42x2yz and 63x3y2z3
Solution:
We know that the numerical coefficients of given numerical are 42 and 63.
Greatest common factor of 42, 63 is 21.
Common literals appearing in given numerical are x, y and z
Smallest power of x in two monomials is 2
Smallest power of y in two monomials is 1
Smallest power of z in two monomials is 1
Monomials of common literals with smallest power is x2yz
∴ The greatest common factor = 21x2yz
5. 12ax2, 6a2x3 and 2a3x5
Solution:
We know that the numerical coefficients of given numerical are 12, 6 and 2
Greatest common factor of 12, 6 and 2 is 2.
Common literals appearing in given numerical are a and x
Smallest power of x in three monomials is 2
Smallest power of a in three monomials is 1
Monomials of common literals with smallest power is ax2
∴ The greatest common factor = 2ax2
6. 9x2, 15x2y3, 6xy2 and 21x2y2
Solution:
We know that the numerical coefficients of given numerical are 9, 15, 16 and 21
Greatest common factor of 9, 15, 16 and 21 is 3.
Common literals appearing in given numerical are x and y
Smallest power of x in four monomials is 1
Smallest power of y in four monomials is 0
Monomials of common literals with smallest power is x
∴ The greatest common factor = 3x
7. 4a2b3, -12a3b, 18a4b3
Solution:
We know that the numerical coefficients of given numerical are 4, -12 and 18.
Greatest common factor of 4, -12 and 18 is 2.
Common literals appearing in given numerical are a and b
Smallest power of a in three monomials is 2
Smallest power of b in three monomials is 1
Monomials of common literals with smallest power is a2b
∴ The greatest common factor = 2a2b
8. 6x2y2, 9xy3, 3x3y2
Solution:
We know that the numerical coefficients of given numerical are 6, 9 and 3
Greatest common factor of 6, 9 and 3 is 3.
Common literals appearing in given numerical are x and y
Smallest power of x in three monomials is 1
Smallest power of y in three monomials is 2
Monomials of common literals with smallest power is xy2
∴ The greatest common factor = 3xy2
9. a2b3, a3b2
Solution:
We know that the numerical coefficients of given numerical are 0
Common literals appearing in given numerical are a and b
Smallest power of a in two monomials = 2
Smallest power of b in two monomials = 2
Monomials of common literals with smallest power is a2b2
∴ The greatest common factor = a2b2
10. 36a2b2c4, 54a5c2, 90a4b2c2
Solution:
We know that the numerical coefficients of given numerical are 36, 54 and 90
Greatest common factor of 36, 54 and 90 is 18.
Common literals appearing in given numerical are a, b and c
Smallest power of a in three monomials is 2
Smallest power of b in three monomials is 0
Smallest power of c in three monomials is 2
Monomials of common literals with smallest power is a2c2
∴ The greatest common factor = 18a2c2
11. x3, -yx2
Solution:
We know that the numerical coefficients of given numerical are 0
Common literals appearing in given numerical are x and y
Smallest power of x in two monomials is 2
Smallest power of y in two monomials is 0
Monomials of common literals with smallest power is x2
∴ The greatest common factor = x2
12. 15a3, -45a2, -150a
Solution:
We know that the numerical coefficients of given numerical are 15, -45 and 150
Greatest common factor of 15, -45 and 150 is 15.
Common literals appearing in given numerical is a
Smallest power of a in three monomials is 1
Monomials of common literals with smallest power is a
∴ The greatest common factor = 15a
13. 2x3y2, 10x2y3, 14xy
Solution:
We know that the numerical coefficients of given numerical are 2, 10 and 14.
Greatest common factor of 2, 10 and 14 is 2.
Common literals appearing in given numerical are x and y
Smallest power of x in three monomials is 1
Smallest power of y in three monomials is 1
Monomials of common literals with smallest power is xy
∴ The greatest common factor = 2xy
14. 14x3y5, 10x5y3, 2x2y2
Solution:
We know that the numerical coefficients of given numerical are 14, 10 and 2.
Greatest common factor of 14, 10 and 2 is 2.
Common literals appearing in given numerical are x and y
Smallest power of x in three monomials is 2
Smallest power of y in three monomials is 2
Monomials of common literals with smallest power is x2y2
∴ The greatest common factor = 2x2y2
Find the greatest common factor of the terms in each of the following expressions:
15. 5a4 + 10a3 – 15a2
Solution:
The greatest common factor of the three terms is 5a2
16. 2xyz + 3x2y + 4y2
Solution:
The greatest common factor of the three terms is y
17. 3a2b2 + 4b2c2 + 12a2b2c2
Solution:
The greatest common factor of the three terms is b2.

EXERCISE 7.2 PAGE NO: 7.5
Factorize the following:
1. 3x – 9
Solution:
The greatest common factor in the given two terms is 3
3x – 9
3 (x – 3)
2. 5x – 15x2
Solution:
The greatest common factor in the given two terms is 5x
5x – 15x2
5x (1 – 3x)
3. 20a12b2 – 15a8b4
Solution:
Greatest common factor in the given two terms is 5a8b2
20a12b2 – 15a8b4
5a8b2 (4a4 – 3b2)
4. 72x6y7 – 96x7y6
Solution:
Greatest common factor in the given two terms is 24x6y6
72x6y7 – 96x7y6
24x6y6 (3y – 4x)
5. 20x3 – 40x2 + 80x
Solution:
Greatest common factor in the given three terms is 20x
20x3 – 40x2 + 80x
20x (x2 – 2x +4)
6. 2x3y2 – 4x2y3 + 8xy4
Solution:
Greatest common factor in the given three terms is 2xy2
2x3y2 – 4x2y3 – 8xy4
2xy2 (x2 – 2xy + 4y2)
7. 10m3n2 + 15m4n – 20m2n3
Solution:
Greatest common factor in the given three terms is 5mn2
10m3n2 + 15m4n – 20m2n3
5m2n (2mn + 3m2 – 4n2)
8. 2a4b4 – 3a3b5 + 4a2b5
Solution:
Greatest common factor in the given three terms is a2b4
2a4b4 – 3a3b5 + 4a2b5
a2b4 (2a2 – 3ab + 4b)
9. 28a2 + 14a2b2 – 21a4
Solution:
Greatest common factor in the given three terms is 7a2
28a2 + 14a2b2 – 21a4
7a(4a + 2b2 – 3a2)
10. a4b – 3a2b2 – 6ab3
Solution:
Greatest common factor in the given three terms is ab
a4b – 3a2b2 – 6ab3
ab (a3 – 3ab – 6b2)
11. 212mn – 3lm2n + 4lmn2
Solution:
Greatest common factor in the given three terms is lmn
2l2mn – 3lm2n + 4lmn2
lmn (2l – 3m + 4n)
12. x4y2 – x2y4 – x4y4
Solution:
Greatest common factor in the given three terms is x2y2
x4y2 – x2y4 – x4y4
x2y2 (x2 – y2 – x2y2)
13. 9x2y + 3axy
Solution:
Greatest common factor in the given three terms is 3xy
9x2y + 3axy
3xy (3x + a)
14. 16m – 4m2
Solution:
Greatest common factor in the given two terms is 4m
16m – 4m2
4m (4 – m)
15. -4a2 + 4ab – 4ca
Solution:
Greatest common factor in the given three terms is – 4a
-4a2 + 4ab – 4ca
-4a (a – b + c)
16. x2yz + xy2z + xyz2
Solution:
Greatest common factor in the given three terms is xyz
x2yz + xy2z + xyz2
xyz (x + y +z)
17. ax2y + bxy2 + cxyz
Solution:
Greatest common factor in the given three terms is xy
ax2y + bxy2 + cxyz
xy (ax + by + cz)

EXERCISE 7.3 PAGE NO: 7.7
Factorize each of the following algebraic expressions:
1. 6x (2x – y) + 7y (2x – y)
Solution:
We have,
6x (2x – y) + 7y (2x – y)
By taking (2x – y) as common we get,
(6x + 7y) (2x – y)
2. 2r (y – x) + s (x – y)
Solution:
We have,
2r (y – x) + s (x – y)
By taking (-1) as common we get,
-2r (x – y) + s (x – y)
By taking (x – y) as common we get,
(x – y) (-2r + s)
(x – y) (s – 2r)
3. 7a (2x – 3) + 3b (2x – 3)
Solution:
We have,
7a (2x – 3) + 3b (2x – 3)
By taking (2x – 3) as common we get,
(7a + 3b) (2x – 3)
4. 9a (6a – 5b) – 12a2 (6a – 5b)
Solution:
We have,
9a (6a – 5b) – 12a2 (6a – 5b)
By taking (6a – 5b) as common we get,
(9a – 12a2) (6a – 5b)
3a(3 – 4a) (6a – 5b)
5. 5 (x – 2y)2 + 3 (x – 2y)
Solution:
We have,
5 (x – 2y)2 + 3 (x – 2y)
By taking (x – 2y) as common we get,
(x – 2y) [5 (x – 2y) + 3]
(x – 2y) (5x – 10y + 3)
6. 16 (2l – 3m)2 – 12 (3m – 2l)
Solution:
We have,
16 (2l – 3m)2 – 12 (3m – 2l)
By taking (-1) as common we get,
16 (2l – 3m)2 + 12 (2l – 3m)
By taking 4(2l – 3m) as common we get,
4(2l – 3m) [4 (2l – 3m) + 3]
4(2l – 3m) (8l – 12m + 3)
7. 3a (x – 2y) – b (x – 2y)
Solution:
We have,
3a (x – 2y) – b (x – 2y)
By taking (x – 2y) as common we get,
(3a – b) (x – 2y)
8. a2 (x + y) + b2 (x + y) + c2 (x + y)
Solution:
We have,
a2 (x + y) + b2 (x + y) + c2 (x + y)
By taking (x + y) as common we get,
(a2 + b2 + c2) (x + y)
9. (x – y)2 + (x – y)
Solution:
We have,
(x – y)2 + (x – y)
By taking (x – y) as common we get,
(x – y) (x – y + 1)
10. 6 (a + 2b) – 4 (a + 2b)2
Solution:
We have,
6 (a + 2b) – 4 (a + 2b)2
By taking (a + 2b) as common we get,
[6 – 4 (a + 2b)] (a + 2b)
(6 – 4a – 8b) (a + 2b)
2(3 – 2a – 4b) (a + 2b)
11. a (x – y) + 2b (y – x) + c (x – y)2
Solution:
We have,
a (x – y) + 2b (y – x) + c (x – y)2
By taking (-1) as common we get,
a (x – y) – 2b (x – y) + c (x – y)2
By taking (x – y) as common we get,
[a – 2b + c(x – y)] (x – y)
(x – y) (a – 2b + cx – cy)
12. -4 (x – 2y)2 + 8 (x – 2y)
Solution:
We have,
-4 (x – 2y)2 + 8 (x – 2y)
By taking 4(x – 2y) as common we get,
[-(x – 2y) + 2] 4(x – 2y)
4(x – 2y) (-x + 2y + 2)
13. x3 (a – 2b) + x2 (a – 2b)
Solution:
We have,
x3 (a – 2b) + x2 (a – 2b)
By taking x2 (a – 2b) as common we get,
(x + 1) [x2 (a – 2b)]
x2 (a – 2b) (x + 1)
14. (2x – 3y) (a + b) + (3x – 2y) (a + b)
Solution:
We have,
(2x – 3y) (a + b) + (3x – 2y) (a + b)
By taking (a + b) as common we get,
(a + b) [(2x – 3y) + (3x – 2y)]
(a + b) [2x -3y + 3x – 2y]
(a + b) [5x – 5y]
(a + b) 5(x – y)
15. 4(x + y) (3a – b) + 6(x + y) (2b – 3a)
Solution:
We have,
4(x + y) (3a – b) + 6(x + y) (2b – 3a)
By taking (x + y) as common we get,
(x + y) [4(3a – b) + 6(2b – 3a)]
(x + y) [12a – 4b + 12b – 18a]
(x + y) [-6a + 8b]
(x + y) 2(-3a + 4b)
(x + y) 2(4b – 3a)

EXERCISE 7.4 PAGE NO: 7.12
Factorize each of the following expressions:
1. qr – pr + qs – ps
Solution:
We have,
qr – pr + qs – ps
By grouping similar terms we get,
qr + qs – pr – ps
q(r + s) –p (r + s)
(q – p) (r + s)
2. p2q – pr2 – pq + r2
Solution:
We have,
p2q – pr2 – pq + r2
By grouping similar terms we get,
p2q – pq – pr2 + r2
pq(p – 1) –r2(p – 1)
(p – 1) (pq – r2)
3. 1 + x + xy + x2y
Solution:
We have,
1 + x + xy + x2y
1 (1 + x) + xy(1 + x)
(1 + x) (1 + xy)
4. ax + ay – bx – by
Solution:
We have,
ax + ay – bx – by
a(x + y) –b (x + y)
(a – b) (x + y)
5. xa2 + xb2 – ya2 – yb2
Solution:
We have,
xa2 + xb2 – ya2 – yb2
x(a2 + b2) –y (a2 + b2)
(x – y) (a2 + b2)
6. x2 + xy + xz + yz
Solution:
We have,
x2 + xy + xz + yz
x (x + y) + z (x + y)
(x + y) (x + z)
7. 2ax + bx + 2ay + by
Solution:
We have,
2ax + bx + 2ay + by
By grouping similar terms we get,
2ax + 2ay + bx + by
2a (x + y) + b (x + y)
(2a + b) (x + y)
8. ab – by – ay + y2
Solution:
We have,
ab – by – ay + y2
By grouping similar terms we get,
Ab – ay – by + y2
a (b – y) – y (b – y)
(a – y) (b – y)
9. axy + bcxy – az – bcz
Solution:
We have,
axy + bcxy – az – bcz
By grouping similar terms we get,
axy – az + bcxy – bcz
a (xy – z) + bc (xy – z)
(a + bc) (xy – z)
10. lm2 – mn2 – lm + n2
Solution:
We have,
lm2 – mn2 – lm + n2
By grouping similar terms we get,
lm2 – lm – mn2 + n2
lm (m – 1) – n2 (m – 1)
(lm – n2) (m – 1)
11. x3 – y2 + x – x2y2
Solution:
We have,
x3 – y2 + x – x2y2
By grouping similar terms we get,
x3 + x – y2 – x2y2
x (1 + x2) – y2 (1 + x2)
(x – y2) (1 + x2)
12. 6xy + 6 – 9y – 4x
Solution:
We have,
6xy + 6 – 9y – 4x
By grouping similar terms we get,
6xy – 4x – 9y + 6
2x (3y – 2) – 3 (3y – 2)
(2x – 3) (3y – 2)
13. x2 – 2ax – 2ab + bx
Solution:
We have,
x2 – 2ax – 2ab + bx
By grouping similar terms we get,
x2 + bx – 2ax – 2ab
x (x + b) – 2a (x + b)
(x – 2a) (x + b)
14. x3 – 2x2y + 3xy2 – 6y3
Solution:
We have,
x3 – 2x2y + 3xy2 – 6y3
By grouping similar terms we get,
x3 + 3xy2 – 2x2y – 6y3
x (x2 + 3y2) – 2y (x2 + 3y2)
(x – 2y) (x2 + 3y2)
15. abx2 + (ay – b) x – y
Solution:
We have,
abx2 + (ay – b) x – y
abx2 – ayx – bx – y
By grouping similar terms we get,
abx2 – bx – ayx – y
bx (ax – 1) + y (ax – 1)
(bx + y) (ax – 1)
16. (ax + by)2 + (bx – ay)2
Solution:
We have,
(ax + by)2 + (bx – ay)2
a2x2 + b2y2 + 2axby + b2x2 + a2y2 – 2axby
a2x2 + b2y2 + b2x2 + a2y2
By grouping similar terms we get,
a2x2 + a2y2 + b2y2 + b2x2
a2 (x2 + y2) + b2 (x2 + y2)
(a2 + b2) (x2 + y2)
17. 16 (a – b)3 – 24 (a – b)2
Solution:
We have,
16(a – b)3 – 24(a – b)2
8 (a – b)2 [2 (a – b) – 3]
8 (a – b)2 (2a – 2b – 3)
18. ab (x2 + 1) + x (a2 + b2)
Solution:
We have,
ab(x2 + 1) + x(a2 + b2)
abx2 + ab + xa2 + xb2
By grouping similar terms we get,
abx2 + xa2 + ab + xb2
ax (bx + a) + b (bx + a)
(ax + b) (bx + a)
19. a2x2 + (ax2 + 1)x + a
Solution:
We have,
a2x2 + (ax2 + 1)x + a
a2x2 + ax3 + x + a
ax2 (a + x) + 1 (x + a)
(x + a) (ax2 + 1)
20. a (a – 2b – c) + 2bc
Solution:
We have,
a (a – 2b – c) + 2bc
a2 – 2ab – ac + 2bc
a (a – 2b) – c (a – 2b)
(a – 2b) (a – c)
21. a (a + b – c) – bc
Solution:
We have,
a (a + b – c) – bc
a2 + ab – ac – bc
a (a + b) – c (a + b)
(a + b) (a – c)
22. x2 – 11xy – x + 11y
Solution:
We have,
x2 – 11xy – x + 11y
By grouping similar terms we get,
x2 – x – 11xy + 11y
x (x – 1) – 11y (x – 1)
(x – 11y) (x – 1)
23. ab – a – b + 1
Solution:
We have,
ab – a – b + 1
a (b – 1) – 1 (b – 1)
(a – 1) (b – 1)
24. x2 + y – xy – x
Solution:
We have,
x2 + y – xy – x
By grouping similar terms we get,
x2 – x + y – xy
x (x – 1) – y (x – 1)
(x – y) (x – 1)

EXERCISE 7.5 PAGE NO: 7.17
Factorize each of the following expressions:
1. 16x2 – 25y2
Solution:
We have,
16x2 – 25y2
(4x)2 – (5y)2
By using the formula (a2 – b2) = (a + b) (a – b) we get,
(4x + 5y) (4x – 5y)
2. 27x2 – 12y2
Solution:
We have,
27x2 – 12y2
By taking 3 as common we get,
3 [(3x)2 – (2y)2]
By using the formula (a2 – b2) = (a-b) (a+b)
3 (3x + 2y) (3x – 2y)
3. 144a2 – 289b2
Solution:
We have,
144a2 – 289b2
(12a)2 – (17b)2
By using the formula (a2 – b2) = (a-b) (a+b)
(12a + 17b) (12a – 17b)
4. 12m2 – 27
Solution:
We have,
12m2 – 27
By taking 3 as common we get,
3 (4m2 – 9)
3 [(2m)2 – 32]
By using the formula (a2 – b2) = (a-b) (a+b)
3 (2m + 3) (2m – 3)
5. 125x2 – 45y2
Solution:
We have,
125x2 – 45y2
By taking 5 as common we get,
5 (25x2 – 9y2)
5 [(5x)2 – (3y)2]
By using the formula (a2 – b2) = (a-b) (a+b)
5 (5x + 3y) (5x – 3y)
6. 144a2 – 169b2
Solution:
We have,
144a2 – 169b2
(12a)2 – (13b)2
By using the formula (a2 – b2) = (a-b) (a+b)
(12a + 13b) (12a – 13b)
7. (2a – b)2 – 16c2
Solution:
We have,
(2a – b)2 – 16c2
(2a – b)2 – (4c)2
By using the formula (a2 – b2) = (a-b) (a+b)
(2a – b + 4c) (2a – b – 4c)
8. (x + 2y)2 – 4 (2x – y)2
Solution:
We have,
(x + 2y)2 – 4 (2x – y)2
(x + 2y)2 – [2 (2x – y)]2
By using the formula (a2 – b2)= (a + b) (a – b) we get,
[(x + 2y) + 2 (2x – y)] [x + 2y – 2 (2x – y)]
(x + 4x + 2y – 2y) (x – 4x + 2y + 2y)
(5x) (4y – 3x)
9. 3a5 – 48a3
Solution:
We have,
3a5 – 48a3
By taking 3 as common we get,
3a3 (a2 – 16)
3a3 (a2 – 42)
By using the formula (a2 – b2) = (a-b) (a+b)
3a3 (a + 4) (a – 4)
10. a4 – 16b4
Solution:
We have,
a4 – 16b4
(a2)2 – (4b2)2
By using the formula (a2 – b2) = (a-b) (a+b)
(a2 + 4b2) (a2 – 4b2)
By using the formula (a2 – b2) = (a-b) (a+b)
(a2 + 4b2) (a + 2b) (a – 2b)
11. x8 – 1
Solution:
We have,
x8 – 1
(x4)2–(1)2
By using the formula (a2 – b2) = (a-b) (a+b)
(x4 + 1) (x4 – 1)
By using the formula (a2 – b2) = (a-b) (a+b)
(x4 + 1) (x2 + 1) (x – 1) (x + 1)
12. 64 – (a + 1)2
Solution:
We have,
64 – (a + 1)2
82 – (a + 1)2
By using the formula (a2 – b2) = (a-b) (a+b)
[8 + (a + 1)] [8 – (a + 1)]
(a + 9) (7 – a)
13. 36l2 – (m + n)2
Solution:
We have,
36l2 – (m + n)2
(6l)2 – (m + n)2
By using the formula (a2 – b2) = (a-b) (a+b)
(6l + m + n) (6l – m – n)
14. 25x4y4 – 1
Solution:
We have,
25x4y4 – 1
(5x2y2)2 – (1)2
By using the formula (a2 – b2) = (a-b) (a+b)
(5x2y2 – 1) (5x2y2 + 1)
15. a4 – 1/b4
Solution:
We have,
a4 – 1/b4
(a2)2 – (1/b2)2
By using the formula (a2 – b2) = (a-b) (a+b)
(a2 + 1/b2) (a2 – 1/b2)
By using the formula (a2 – b2) = (a-b) (a+b)
(a2 + 1/b2) (a – 1/b) (a + 1/b)
16. x3 – 144x
Solution:
We have,
x3 – 144x
x [x2 – (12)2]
By using the formula (a2 – b2) = (a-b) (a+b)
x (x + 12) (x – 12)
17. (x – 4y)2 – 625
Solution:
We have,
(x – 4y)2 – 625
(x – 4y)2 – (25)2
By using the formula (a2 – b2) = (a-b) (a+b)
(x – 4y + 25) (x – 4y – 25)
18. 9 (a – b)2 – 100 (x – y)2
Solution:
We have,
9 (a – b)2 – 100 (x – y)2
[3 (a – b)]2 – [10 (x – y)]2
By using the formula (a2 – b2) = (a-b) (a+b)
[3 (a – b) + 10 (x + y)] [3 (a – b) – 10 (x – y)] [3a – 3b + 10x – 10y] [3a – 3b – 10x + 10y]
19. (3 + 2a)2 – 25a2
Solution:
We have,
(3 + 2a)2 – 25a2
(3 + 2a)2 – (5a)2
By using the formula (a2 – b2) = (a-b) (a+b)
(3 + 2a + 5a) (3 + 2a – 5a)
(3 + 7a) (3 – 3a)
(3 + 7a) 3(1 – a)
20. (x + y)2 – (a – b)2
Solution:
We have,
(x + y)2 – (a – b)2
By using the formula (a2 – b2) = (a-b) (a+b)
[(x + y) + (a – b)] [(x + y) – (a – b)]
(x + y + a – b) (x + y – a + b)
21. 1/16x2y2 – 4/49y2z2
Solution:
We have,
1/16x2y2 – 4/49y2z2
(1/4xy)2 – (2/7yz)2
By using the formula (a2 – b2) = (a-b) (a+b)
(xy/4 + 2yz/7) (xy/4 – 2yz/7)
y2 (x/4 + 2/7z) (x/4 – 2/7z)
22. 75a3b2 – 108ab4
Solution:
We have,
75a3b2 – 108ab4
3ab2 (25a2 – 36b2)
3ab2 [(5a)2 – (6b)2]
By using the formula (a2 – b2) = (a-b) (a+b)
3ab2 (5a + 6b) (5a – 6b)
23. x5 – 16x3
Solution:
We have,
x5 – 16x3
x3 (x2 – 16)
x3 (x2 – 42)
By using the formula (a2 – b2) = (a-b) (a+b)
x3 (x + 4) (x – 4)
24. 50/x2 – 2x2/81
Solution:
We have,
50/x2 – 2x2/81
2 (25/x2 – x2/81)
2 [(5/x)2 – (x/9)2]
By using the formula (a2 – b2) = (a-b) (a+b)
2 (5/x+ x/9) (5/x – x/9)
25. 256x3 – 81x
Solution:
We have,
256x3 – 81x
x (256x4 – 81)
x [(16x2)2 – 92]
By using the formula (a2 – b2) = (a-b) (a+b)
x (4x + 3) (4x – 3) (16x2 + 9)
26. a4 – (2b + c)4
Solution:
We have,
a4 – (2b + c)4
(a2)2 – [(2b + c)2]2
By using the formula (a2 – b2) = (a-b) (a+b)
[a2 + (2b + c)2] [a2 – (2b + c)2]
By using the formula (a2 – b2) = (a-b) (a+b)
[a2 + (2b + c)2] [a + 2b + c] [a – 2b – c]
27. (3x + 4y)4 – x4
Solution:
We have,
(3x + 4y)4 – x4
[(3x + 4y)2]2 – (x2)2
By using the formula (a2 – b2) = (a-b) (a+b)
[(3x + 4y)2 + x2] [(3x + 4y)2 – x2] [(3x + 4y)2 + x2] [3x + 4y + x] [3x + 4y – x] [(3x + 4y)2 + x2] [4x + 4y] [2x + 4y] [(3x + 4y)2 + x2] 8[x + 2y] [x + y]
28. p2q2 – p4q4
Solution:
We have,
p2q2 – p4q4
(pq)2 – (p2q2)2
By using the formula (a2 – b2) = (a-b) (a+b)
(pq + p2q2) (pq – p2q2)
p2q2 (1 + pq) (1 – pq)
29. 3x3y – 24xy3
Solution:
We have,
3x3y – 24xy3
3xy (x2 – 8y2)
3xy [x2 – (8y)2]
By using the formula (a2 – b2) = (a-b) (a+b)
(3xy) (x + 8y) (x – 8y)
30. a4b4 – 16c4
Solution:
We have,
a4b4 – 16c4
(a2b2)2 – (4c2)2
By using the formula (a2 – b2) = (a-b) (a+b)
(a2b2 + 4c2) (a2b2 – 4c2)
By using the formula (a2 – b2) = (a-b) (a+b)
(a2b2 + 4c2) (ab + 2c) (ab – 2c)
31. x4 – 625
Solution:
We have,
x4 – 625
(x2)2 – (25)2
By using the formula (a2 – b2) = (a-b) (a+b)
(x2 + 25) (x2 – 25)
(x2 + 25) (x2 – 52)
By using the formula (a2 – b2) = (a-b) (a+b)
(x2 + 25) (x + 5) (x – 5)
32. x4 – 1
Solution:
We have,
x4 – 1
(x2)2 – (1)2
By using the formula (a2 – b2) = (a-b) (a+b)
(x2 + 1) (x2 – 1)
By using the formula (a2 – b2) = (a-b) (a+b)
(x2 + 1) (x + 1) (x – 1)
33. 49(a – b)2 – 25(a + b)2
Solution:
We have,
49(a – b)2 – 25(a + b)2
[7 (a – b)]2 – [5 (a + b)]2
By using the formula (a2 – b2) = (a-b) (a+b)
[7 (a – b) + 5 (a + b)] [7 (a – b) – 5 (a + b)]
(7a – 7b + 5a + 5b) (7a – 7b – 5a – 5b)
(12a – 2b) (2a – 12b)
2 (6a – b) 2 (a – 6b)
4 (6a – b) (a – 6b)
34. x – y – x2 + y2
Solution:
We have,
x – y – x2 + y2
x – y – (x2 – y2)
By using the formula (a2 – b2) = (a-b) (a+b)
x – y – (x + y) (x – y)
(x – y) (1 – x – y)
35. 16(2x – 1)2 – 25y2
Solution:
We have,
16(2x – 1)2 – 25y2
[4 (2x – 1)]2 – (5y)2
By using the formula (a2 – b2) = (a-b) (a+b)
(8x + 5y – 4) (8x – 5y – 4)
36. 4(xy + 1)2 – 9(x – 1)2
Solution:
We have,
4(xy + 1)2 – 9(x – 1)2
[2x (xy + 1)]2 – [3 (x – 1)]2
By using the formula (a2 – b2) = (a-b) (a+b)
(2xy + 2 + 3x – 3) (2xy + 2 – 3x + 3)
(2xy + 3x – 1) (2xy – 3x + 5)
37. (2x + 1)2 – 9x4
Solution:
We have,
(2x + 1)2 – 9x4
(2x + 1)2 – (3x2)2
By using the formula (a2 – b2) = (a-b) (a+b)
(2x + 1 + 3x2) (2x + 1 – 3x2)
(3x2 + 2x + 1) (-3x2 + 2x + 1)
38. x4 – (2y – 3z)2
Solution:
We have,
x4 – (2y – 3z)2
(x2)2 – (2y – 3z)2
By using the formula (a2 – b2) = (a-b) (a+b)
(x2 + 2y – 3z) (x2 – 2y + 3z)
39. a4 – b2 + a – b
Solution:
We have,
a4 – b2 + a – b
By using the formula (a2 – b2) = (a-b) (a+b)
(a + b) (a – b) + (a – b)
(a – b) (a + b + 1)
40. 16a4 – b4
Solution:
We have,
16a4 – b4
(4a2)2 – (b2)2
(4a2 + b2) (4a2 – b2)
By using the formula (a2 – b2) = (a-b) (a+b)
(4a2 + b2) (2a + b) (2a – b)
41. a4 – 16(b – c)4
Solution:
We have,
a4 – 16(b – c)4
(a2)2 – [4 (b – c)2]
By using the formula (a2 – b2) = (a-b) (a+b)
[a2 + 4 (b – c)2] [a2 – 4 (b – c)2]
By using the formula (a2 – b2) = (a-b) (a+b)
[a2 + 4 (b – c)2] [(a + 2b – 2c) (a – 2b + 2c)]
42. 2a4 – 32a
Solution:
We have,
2a4 – 32a
2a (a4 – 16)
2a [(a)2 – (4)2]
By using the formula (a2 – b2) = (a-b) (a+b)
2a (a2 + 4) (a2 – 4)
2a (a2 + 4) (a2 – 22)
By using the formula (a2 – b2) = (a-b) (a+b)
2a (a2 + 4) (a + 2) (a – 2)
43. a4b4 – 81c4
Solution:
We have,
a4b4 – 81c4
(a2b2)2 – (9c2)2
By using the formula (a2 – b2) = (a-b) (a+b)
(a2b2 + 9c2) (a2b2 – 9c2)
By using the formula (a2 – b2) = (a-b) (a+b)
(a2b2 + 9c2) (ab + 3c) (ab – 3c)
44. xy9 – yx9
Solution:
We have,
xy9 – yx9
-xy (x8 – y8)
-xy [(x4)2 – (y4)2]
By using the formula (a2 – b2) = (a-b) (a+b)
-xy (x4 + y4) (x4 – y4)
By using the formula (a2 – b2) = (a-b) (a+b)
-xy (x4 + y4) (x2 + y2) (x2 – y2)
By using the formula (a2 – b2) = (a-b) (a+b)
-xy (x4 + y4) (x2 + y2) (x + y) (x – y)
45. x3 – x
Solution:
We have,
x3 – x
x (x2 – 1)
By using the formula (a2 – b2) = (a-b) (a+b)
x (x + 1) (x – 1)
46. 18a2x2 – 32
Solution:
We have,
18a2x2 – 32
2 [(3ax)2 – (4)2]
By using the formula (a2 – b2) = (a-b) (a+b)
2 (3ax + 4) (3ax – 4)

EXERCISE 7.6 PAGE NO: 7.22
Factorize each of the following algebraic expressions:
1. 4x2 + 12xy + 9y2
Solution:
We have,
4x2 + 12xy + 9y2
By using the formula (x + y)2 = x2 + y2 + 2xy
(2x)2 + (3y)2 + 2 (2x) (3y)
(2x + 3y)2
(2x + 3y) (2x + 3y)
2. 9a2 – 24ab + 16b2
Solution:
We have,
9a2 – 24ab + 16b2
By using the formula (x – y)2 = x2 + y2 – 2xy
Here x = 3a, y = 4b So,
(3a)2 + (4b)2 – 2 (3a) (4b)
(3a – 4b)2
(3a – 4b) (3a – 4b)
3. p2q2 – 6pqr + 9r2
Solution:
We have,
p2q2 – 6pqr + 9r2
By using the formula (a – b)2 = a2 + b2 – 2ab
(pq)2 + (3r)2 – 2 (pq) (3r)
(pq – 3r)2
(pq – 3r) (pq – 3r)
4. 36a2 + 36a + 9
Solution:
We have,
36a2 + 36a + 9
9 (4a2 + 4a + 1)
9[(2a)2 + 2 (2a) + 11]
9 (2a + 1)2
9 (2a + 1) (2a + 1)
5. a2 + 2ab + b2 – 16
Solution:
We have,
a2 + 2ab + b2 – 16
By using the formula (a – b)2 = a2 + b2 – 2ab
(a + b)2 – 42
By using the formula (a2 – b2) = (a+b) (a-b)
(a + b + 4) (a + b – 4)
6. 9z2 – x2 + 4xy – 4y2
Solution:
We have,
9z2 – x2 + 4xy – 4y2
(3z)2 – [x2 – 2 (x) (2y) + (2y)2]
By using the formula (a – b)2 = a2 + b2 – 2ab
(3z)2 – (x – 2y)2
By using the formula (a2 – b2) = (a+b) (a-b)
[(x – 2y) + 3z] [–x + 2y + 3z)]
7. 9a2 – 24a2b2 + 16b4 – 256
Solution:
We have,
9a2 – 24a2b2 + 16b4 – 256
(3a2)2 – 2 (4a2) (3b2) + (4b2)2 – (16)2
By using the formula (a – b)2 = a2 + b2 – 2ab
(3a2 – 4b2)2 – (16)2
By using the formula (a2 – b2) = (a+b) (a-b)
(3a2 – 4b2 + 16) (3a2 – 4b2 – 16)
8. 16 – a6 + 4a3b3 – 4b6
Solution:
We have,
16 – a6 + 4a3b3 – 4b6
42 – [(a3)2 – 2 (a3) (2b3) + (2b3)2]
By using the formula (a – b)2 = a2 + b2 – 2ab
42 – (a3 – 2b3)2
By using the formula (a2 – b2) = (a+b) (a-b)
[4 + (a3 – 2b3)] [4 – (a3 – 2b3)]
9. a2 – 2ab + b2 – c2
Solution:
We have,
a2 – 2ab + b2 – c2
By using the formula (a – b)2 = a2 + b2 – 2ab
(a – b)2 – c2
By using the formula (a2 – b2) = (a+b) (a-b)
(a – b + c) (a – b – c)
10. x2 + 2x + 1 – 9y2
Solution:
We have,
x2 + 2x + 1 – 9y2
By using the formula (a – b)2 = a2 + b2 – 2ab
(x + 1)2 – (3y)2
By using the formula (a2 – b2) = (a+b) (a-b)
(x + 3y + 1) (x – 3y + 1)
11. a2 + 4ab + 3b2
Solution:
We have,
a2 + 4ab + 3b2
By using factors for 3 i.e., 3 and 1
a2 + ab + 3ab + 3b2
By grouping we get,
a (a + b) + 3b (a + b)
(a + 3b) (a + b)
12. 96 – 4x – x2
Solution:
We have,
96 – 4x – x2
-x2 – 4x + 96
By using factors for 96 i.e., 12 and 8
-x2 – 12x + 8x + 96
By grouping we get,
-x (x + 12) + 8 (x + 12)
(x + 12) (-x + 8)
13. a4 + 3a2 + 4
Solution:
We have,
a4 + 3a2 + 4
(a2)2 + (a2)2 + 2 (2a2) + 4 – a2
(a2 + 2)2 + (-a2)
By using the formula (a2 – b2) = (a+b) (a-b)
(a2 + 2 + a) (a2 + 2 – a)
(a2 + a + 2) (a2 – a + 2)
14. 4x4 + 1
Solution:
We have,
4x4 + 1
(2x2)2 + 1 + 4x2 – 4x2
(2x2 + 1)2 – 4x2
By using the formula (a2 – b2) = (a+b) (a-b)
(2x2 + 1 + 2x) (2x2 + 1 – 2x)
(2x2 + 2x + 1) (2x2 – 2x + 1)
15. 4x4 + y4
Solution:
We have,
4x4 + y4
(2x2)2 + (y2)2 + 4x2y2 – 4x2y2
(2x2 + y2)2 – 4x2y2
By using the formula (a2 – b2) = (a+b) (a-b)
(2x2 + y2 + 2xy) (2x2 + y2 – 2xy)
16. (x + 2)4 – 6(x + 2) + 9
Solution:
We have,
(x + 2)4 – 6(x + 2) + 9
(x2 + 22)2 – 6x – 12 + 9
(x2 + 22 + 2(2)(x)) – 6x – 12 + 9
x2 + 4 + 4x – 6x – 12 + 9
x2 – 2x + 1
By using the formula (a – b)2 = a2 + b2 – 2ab
(x – 1)2
17. 25 – p2 – q2 – 2pq
Solution:
We have,
25 – p2 – q2 – 2pq
25 – (p2 + q2 + 2pq)
(5)2 – (p + q)2
By using the formula (a2 – b2) = (a+b) (a-b)
(5 + p + q) (5 –p – q)
-(p + q + 5) (p + q – 5)
18. x2 + 9y2 – 6xy – 25a2
Solution:
We have,
x2 + 9y2 – 6xy – 25a2
(x – 3y)2 – (5a)2
By using the formula (a2 – b2) = (a+b) (a-b)
(x – 3y + 5a) (x – 3y – 5a)
19. 49 – a2 + 8ab – 16b2
Solution:
We have,
49 – a2 + 8ab – 16b2
49 – (a2 – 8ab + 16b2)
49 – (a – 4b)2
By using the formula (a2 – b2) = (a + b) (a – b)
(7 + a – 4b) (7 – a + 4b)
-(a – 4b + 7) (a – 4b – 7)
20. a2 – 8ab + 16b2 – 25c2
Solution:
We have,
a2 – 8ab + 16b2 – 25c2
(a – 4b)2– (5c)2
By using the formula (a2 – b2) = (a+b) (a-b)
(a – 4b + 5c) (a – 4b – 5c)
21. x2 – y2 + 6y – 9
Solution:
We have,
x2 – y2 + 6y – 9
x2 + 6y – (y2 – 6y + 9)
x2 – (y – 3)2
By using the formula (a2 – b2) = (a+b) (a-b)
(x + y – 3) (x – y + 3)
22. 25x2 – 10x + 1 – 36y2
Solution:
We have,
25x2 – 10x + 1 – 36y2
(5x)2 – 2 (5x) + 1 – (6y)2
(5x – 1)2 – (6y)2
By using the formula (a2 – b2) = (a+b) (a-b)
(5x – 6y – 1) (5x + 6y – 1)
23. a2 – b2 + 2bc – c2
Solution:
We have,
a2 – b2 + 2bc – c2
a2 – (b2 – 2bc + c2)
a2 – (b – c)2
By using the formula (a2 – b2) = (a+b) (a-b)
(a + b – c) (a – b + c)
24. a4 + 2b + b2 – c2
Solution:
We have,
a4 + 2b + b2 – c2
(a + b)2 – c2
By using the formula (a2 – b2) = (a+b) (a-b)
(a + b + c) (a + b – c)
25. 49 – x2 – y2 + 2xy
Solution:
We have,
49 – x2 – y2 + 2xy
49 – (x2 + y2 – 2xy)
72 – (x – y)2
By using the formula (a2 – b2) = (a+b) (a-b)
[7 + (x – y)] [7 – x + y]
(x – y + 7) (y – x + 7)
26. a2 + 4b2 – 4ab – 4c2
Solution:
We have,
a2 + 4b2 – 4ab – 4c2
a2 – 2 (a) (2b) + (2b)2 – (2c)2
(a – 2b)2 – (2c)2
By using the formula (a2 – b2) = (a+b) (a-b)
(a – 2b + 2c) (a – 2b – 2c)
27. x2 – y2 – 4xz + 4z2
Solution:
We have,
x2 – y2 – 4xz + 4z2
x2 – 2 (x) (2z) + (2z)2 – y2
As (a-b)2 = a2 + b2 – 2ab
(x – 2z)2 – y2
By using the formula (a2 – b2) = (a+b) (a-b)
(x + y – 2z) (x – y – 2z)

EXERCISE 7.7 PAGE NO: 7.27
Factorize each of the following algebraic expressions:
1. x2 + 12x – 45
Solution:
We have,
x2 + 12x – 45
To factorize the given expression we have to find two numbers p and q such that p+q = 12 and pq = -45
So we can replace 12x by 15x – 3x
-45 by 15 × 3
x2 + 12x – 45 = x2 + 15x – 3x – 45
= x (x + 15) – 3 (x + 15)
= (x – 3) (x + 15)
2. 40 + 3x – x2
Solution:
We have,
40 + 3x – x2
-(x2 – 3x – 40)
By considering, p+q = -3 and pq = -40
So we can replace -3x by 5x – 8x
-40 by 5 × -8
-(x2 – 3x – 40) = x2 + 5x – 8x – 40
= -x (x + 5) – 8 (x + 5)
= -(x – 8) (x + 5)
= (-x + 8) (x + 5)
3. a2 + 3a – 88
Solution:
We have,
a2 + 3a – 88
By considering, p+q = 3 and pq = -88
So we can replace 3a by 11a – 8a
-40 by -11 × 8
a2 + 3a – 88 = a2 + 11a – 8a – 88
= a (a + 11) – 8 (a + 11)
= (a – 8) (a + 11)
4. a2 – 14a – 51
Solution:
We have,
a2 – 14a – 51
By considering, p+q = -14 and pq = -51
So we can replace -14a by 3a – 17a
-51 by -17 × 3
a2 – 14a – 51 = a2 + 3a – 17a – 51
= a (a + 3) – 17 (a + 3)
= (a – 17) (a + 3)
5. x2 + 14x + 45
Solution:
We have,
x2 + 14x + 45
By considering, p+q = 14 and pq = 45
So we can replace 14x by 5x + 9x
45 by 5 × 9
x2 + 14x + 45 = x2 + 5x + 9x + 45
= x (x + 5) – 9 (x + 5)
= (x + 9) (x + 5)
6. x2 – 22x + 120
Solution:
We have,
x2 – 22x + 120
By considering, p+q = -22 and pq = 120
So we can replace -22x by -12x -10x
120 by -12 × -10
x2 – 22x + 120 = x2 – 12x – 10x + 120
= x (x – 12) – 10 (x – 12)
= (x – 10) (x – 12)
7. x2 – 11x – 42
Solution:
We have,
x2 – 11x – 42
By considering, p+q = -11 and pq = -42
So we can replace -11x by 3x -14x
-42 by 3 × -14
x2 – 11x – 42 = x2 + 3x – 14x – 42
= x (x + 3) – 14 (x + 3)
= (x – 14) (x + 3)
8. a2 + 2a – 3
Solution:
We have,
a2 + 2a – 3
By considering, p+q = 2 and pq = -3
So we can replace 2a by 3a -a
-3 by 3 × -1
a2 + 2a – 3 = a2 + 3a – a – 3
= a (a + 3) – 1 (a + 3)
= (a – 1) (a + 3)
9. a2 + 14a + 48
Solution:
We have,
a2 + 14a + 48
By considering, p+q = 14 and pq = 48
So we can replace 14a by 8a + 6a
48 by 8 × 6
a2 + 14a + 48 = a2 + 8a + 6a + 48
= a (a + 8) + 6 (a + 8)
= (a + 6) (a + 8)
10. x2 – 4x – 21
Solution:
We have,
x2 – 4x – 21
By considering, p+q = -4 and pq = -21
So we can replace -4x by 3x – 7x
-21 by 3 × -7
x2 + 4x – 21 = x2 + 3x – 7x – 21
= x (x + 3) – 7 (x + 3)
= (x – 7) (x + 3)
11. y2 + 5y – 36
Solution:
We have,
y2 + 5y – 36
By considering, p+q = 5 and pq = -36
So we can replace 5y by 9y – 4y
-36 by 9 × -4
y2 + 5y – 36 = y2 + 9y – 4y – 36
= y (y + 9) – 4 (y + 9)
= (y – 4) (y + 9)
12. (a2 – 5a)2 – 36
Solution:
We have,
(a2 – 5a)2 – 36
(a2 – 5a)2 – 62
By using the formula (a2 – b2) = (a+b) (a-b)
(a2 – 5a)2 – 62 = (a2 – 5a + 6) (a2 – 5a – 6)
So now we shall factorize the expression (a2 – 5a + 6)
By considering, p+q = -5 and pq = 6
So we can replace -5a by a -6a
6 by 1 × -6
a2 -5a – 6 = a2 + a – 6a – 6
= a (a + 1) -6(a + 1)
= (a – 6) (a + 1)
So now we shall factorize the expression (a2 – 5a + 6)
By considering, p+q = -5 and pq = -6
So we can replace -5a by -2a -3a
6 by -2 × -3
a2 -5a + 6 = a2 – 2a – 3a + 6
= a (a – 2) -3 (a – 2)
= (a – 3) (a – 2)
∴ (a2 – 5a)2 – 36 = (a2 – 5a + 6) (a2 – 5a – 6)
= (a + 1) (a – 6) (a – 2) (a – 3)
13. (a + 7) (a – 10) + 16
Solution:
We have,
(a + 7) (a – 10) + 16
a2 – 10a + 7a – 70 + 16
a2 – 3a – 54
By considering, p+q = -3 and pq = -54
So we can replace -3a by 6a – 9a
-54 by 6 × -9
a2 – 3a – 54 = a2 + 6a – 9a – 54
= a (a + 6) -9 (a + 6)
= (a – 9) (a + 6)

EXERCISE 7.8 PAGE NO: 7.30
Resolve each of the following quadratic trinomials into factors:
1. 2x2 + 5x + 3
Solution:
We have,
2x2 + 5x + 3
The coefficient of x2 is 2
The coefficient of x is 5
Constant term is 3
We shall split up the center term i.e., 5 into two parts such that their sum p+q is 5 and product pq = 2 × 3 is 6
So, we express the middle term 5x as 2x + 3x
2x2 + 5x + 3 = 2x2 + 2x + 3x + 3
= 2x (x + 1) + 3 (x + 1)
= (2x + 3) (x + 1)
2. 2x2 – 3x – 2
Solution:
We have,
2x2 – 3x – 2
The coefficient of x2 is 2
The coefficient of x is -3
Constant term is -2
So, we express the middle term -3x as -4x + x
2x2 – 3x – 2 = 2x2 – 4x + x – 2
= 2x (x – 2) + 1 (x – 2)
= (x – 2) (2x + 1)
3. 3x2 + 10x + 3
Solution:
We have,
3x2 + 10x + 3
The coefficient of x2 is 3
The coefficient of x is 10
Constant term is 3
So, we express the middle term 10x as 9x + x
3x2 + 10x + 3 = 3x2 + 9x + x + 3
= 3x (x + 3) + 1 (x + 3)
= (3x + 1) (x + 3)
4. 7x – 6 – 2x2
Solution:
We have,
7x – 6 – 2x2
– 2x2 + 7x – 6
2x2 – 7x + 6
The coefficient of x2 is 2
The coefficient of x is -7
Constant term is 6
So, we express the middle term -7x as -4x – 3x
2x2 – 7x + 6 = 2x2 – 4x – 3x + 6
= 2x (x – 2) – 3 (x – 2)
= (x – 2) (2x – 3)
5. 7x2 – 19x – 6
Solution:
We have,
7x2 – 19x – 6
The coefficient of x2 is 7
The coefficient of x is -19
Constant term is -6
So, we express the middle term -19x as 2x – 21x
7x2 – 19x – 6 = 7x2 + 2x – 21x – 6
= x (7x + 2) – 3 (7x + 2)
= (7x + 2) (x – 3)
6. 28 – 31x – 5x2
Solution:
We have,
28 – 31x – 5x2
– 5x2 -31x + 28
5x2 + 31x – 28
The coefficient of x2 is 5
The coefficient of x is 31
Constant term is -28
So, we express the middle term 31x as -4x + 35x
5x2 + 31x – 28 = 5x2 – 4x + 35x – 28
= x (5x – 4) + 7 (5x – 4)
= (x + 7) (5x – 4)
7. 3 + 23y – 8y2
Solution:
We have,
3 + 23y – 8y2
– 8y2 + 23y + 3
8y2 – 23y – 3
The coefficient of y2 is 8
The coefficient of y is -23
Constant term is -3
So, we express the middle term -23y as -24y + y
8y2 – 23y – 3 = 8y2 – 24y + y – 3
= 8y (y – 3) + 1 (y – 3)
= (8y + 1) (y – 3)
8. 11x2 – 54x + 63
Solution:
We have,
11x2 – 54x + 63
The coefficient of x2 is 11
The coefficient of x is -54
Constant term is 63
So, we express the middle term -54x as -33x – 21x
11x2 – 54x + 63 = 11x2 – 33x – 21x – 6
= 11x (x – 3) – 21 (x – 3)
= (11x – 21) (x – 3)
9. 7x – 6x2 + 20
Solution:
We have,
7x – 6x2 + 20
– 6x2 + 7x + 20
6x2 – 7x – 20
The coefficient of x2 is 6
The coefficient of x is -7
Constant term is -20
So, we express the middle term -7x as -15x + 8x
6x2 – 7x – 20 = 6x2 – 15x + 8x – 20
= 3x (2x – 5) + 4 (2x – 5)
= (3x + 4) (2x – 5)
10. 3x2 + 22x + 35
Solution:
We have,
3x2 + 22x + 35
The coefficient of x2 is 3
The coefficient of x is 22
Constant term is 35
So, we express the middle term 22x as 15x + 7x
3x2 + 22x + 35 = 3x2 + 15x + 7x + 35
= 3x (x + 5) + 7 (x + 5)
= (3x + 7) (x+ 5)
11. 12x2 – 17xy + 6y2
Solution:
We have,
12x2 – 17xy + 6y2
The coefficient of x2 is 12
The coefficient of x is -17y
Constant term is 6y2
So, we express the middle term -17xy as -9xy – 8xy
12x2 -17xy+ 6y2 = 12x2 – 9xy – 8xy + 6y2
= 3x (4x – 3y) – 2y (4x – 3y)
= (3x – 2y) (4x – 3y)
12. 6x2 – 5xy – 6y2
Solution:
We have,
6x2 – 5xy – 6y2
The coefficient of x2 is 6
The coefficient of x is -5y
Constant term is -6y2
So, we express the middle term -5xy as 4xy – 9xy
6x2 -5xy- 6y2 = 6x2 + 4xy – 9xy – 6y2
= 2x (3x + 2y) -3y (3x + 2y)
= (2x – 3y) (3x + 2y)
13. 6x2 – 13xy + 2y2
Solution:
We have,
6x2 – 13xy + 2y2
The coefficient of x2 is 6
The coefficient of x is -13y
Constant term is 2y2
So, we express the middle term -13xy as -12xy – xy
6x2 -13xy+ 2y2 = 6x2 – 12xy – xy + 2y2
= 6x (x – 2y) – y (x – 2y)
= (6x – y) (x – 2y)
14. 14x2 + 11xy – 15y2
Solution:
We have,
14x2 + 11xy – 15y2
The coefficient of x2 is 14
The coefficient of x is 11y
Constant term is -15y2
So, we express the middle term 11xy as 21xy – 10xy
14x2 + 11xy- 15y2 = 14x2 + 21xy – 10xy – 15y2
= 2x (7x – 5y) + 3y (7x – 5y)
= (2x + 3y) (7x – 5y)
15. 6a2 + 17ab – 3b2
Solution:
We have,
6a2 + 17ab – 3b2
The coefficient of a2 is 6
The coefficient of a is 17b
Constant term is -3b2
So, we express the middle term 17ab as 18ab – ab
6a2 +17ab– 3b2 = 6a2 + 18ab – ab – 3b2
= 6a (a + 3b) – b (a + 3b)
= (6a – b) (a + 3b)
16. 36a2 + 12abc – 15b2c2
Solution:
We have,
36a2 + 12abc – 15b2c2
The coefficient of a2 is 36
The coefficient of a is 12bc
Constant term is -15b2c2
So, we express the middle term 12abc as 30abc – 18abc
36a2 –12abc– 15b2c2 = 36a2 + 30abc – 18abc – 15b2c2
= 6a (6a + 5bc) – 3bc (6a + 5bc)
= (6a + 5bc) (6a – 3bc)
= (6a + 5bc) 3(2a – bc)
17. 15x2 – 16xyz – 15y2z2
Solution:
We have,
15x2 – 16xyz – 15y2z2
The coefficient of x2 is 15
The coefficient of x is -16yz
Constant term is -15y2z2
So, we express the middle term -16xyz as -25xyz + 9xyz
15x2 -16xyz- 15y2z2 = 15x2 – 25yz + 9yz – 15y2z2
= 5x (3x – 5yz) + 3yz (3x – 5yz)
= (5x + 3yz) (3x – 5yz)
18. (x – 2y)2 – 5 (x – 2y) + 6
Solution:
We have,
(x – 2y)2 – 5 (x – 2y) + 6
The coefficient of (x-2y)2 is 1
The coefficient of (x-2y) is -5
Constant term is 6
So, we express the middle term -5(x – 2y) as -2(x – 2y) -3(x – 2y)
(x – 2y)2 – 5 (x – 2y) + 6 = (x – 2y)2 – 2 (x – 2y) – 3 (x – 2y) + 6
= (x – 2y – 2) (x – 2y – 3)
19. (2a – b)2 + 2 (2a – b) – 8
Solution:
We have,
(2a – b)2 + 2 (2a – b) – 8
The coefficient of (2a-b)2 is 1
The coefficient of (2a-b) is 2
Constant term is -8
So, we express the middle term 2(2a – b) as 4 (2a –b) – 2 (2a – b)
(2a – b)2 + 2 (2a – b) – 8 = (2a – b)2 + 4 (2a – b) – 2 (2a – b) – 8
= (2a – b) (2a – b + 4) – 2 (2a – b + 4)
= (2a – b + 4) (2a – b – 2)

EXERCISE 7.9 PAGE NO: 7.32
Factorize each of the following quadratic polynomials by using the method of completing the square:
1. p2 + 6p + 8
Solution:
We have,
p2 + 6p + 8
Coefficient of p2 is unity. So, we add and subtract square of half of coefficient of p.
p2 + 6p + 8 = p2 + 6p + 32 – 32 + 8 (Adding and subtracting 32)
= (p + 3)2 – 12 (By completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= (p + 3 – 1) (p + 3 + 1)
= (p + 2) (p + 4)
2. q2 – 10q + 21
Solution:
We have,
q2 – 10q + 21
Coefficient of q2 is unity. So, we add and subtract square of half of coefficient of q.
q2 – 10q + 21 = q2 – 10q+ 52 – 52 + 21 (Adding and subtracting 52)
= (q – 5)2 – 22 (By completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= (q – 5 – 2) (q – 5 + 2)
= (q – 3) (q – 7)
3. 4y2 + 12y + 5
Solution:
We have,
4y2 + 12y + 5
4(y2 + 3y + 5/4)
Coefficient of y2 is unity. So, we add and subtract square of half of coefficient of y.
4(y2 + 3y + 5/4) = 4 [y2 + 3y + (3/2)2 – (3/2)2 + 5/4] (Adding and subtracting (3/2)2)
= 4 [(y + 3/2)2 – 12] (Completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= 4 (y + 3/2 + 1) (y + 3/2 – 1)
= 4 (y + 1/2) (y + 5/2) (by taking LCM)
= 4 [(2y + 1)/2] [(2y + 5)/2]
= (2y + 1) (2y + 5)
4. p2 + 6p – 16
Solution:
We have,
p2 + 6p – 16
Coefficient of p2 is unity. So, we add and subtract square of half of coefficient of p.
p2 + 6p – 16 = p2 + 6p + 32 – 32 – 16 (Adding and subtracting 32)
= (p + 3)2 – 52 (Completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= (p + 3 + 5) (p + 3 – 5)
= (p + 8) (p – 2)
5. x2 + 12x + 20
Solution:
We have,
x2 + 12x + 20
Coefficient of x2 is unity. So, we add and subtract square of half of coefficient of x.
x2 + 12x + 20 = x2 + 12x + 62 – 62 + 20 (Adding and subtracting 62)
= (x + 6)2 – 42 (Completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= (x + 6 + 4) (x + 6 – 4)
= (x + 2) (x + 10)
6. a2 – 14a – 51
Solution:
We have,
a2 – 14a – 51
Coefficient of a2 is unity. So, we add and subtract square of half of coefficient of a.
a2 – 14a – 51 = a2 – 14a + 72 – 72 – 51 (Adding and subtracting 72)
= (a – 7)2 – 102 (Completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= (a – 7 + 10) (9 – 7 – 10)
= (a – 17) (a + 3)
7. a2 + 2a – 3
Solution:
We have,
a2 + 2a – 3
Coefficient of a2 is unity. So, we add and subtract square of half of coefficient of a.
a2 + 2a – 3 = a2 + 2a + 12 – 12 – 3 (Adding and subtracting 12)
= (a + 1)2 – 22 (Completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= (a + 1 + 2) (a + 1 – 2)
= (a + 3) (a – 1)
8. 4x2 – 12x + 5
Solution:
We have,
4x2 – 12x + 5
4(x2 – 3x + 5/4)
Coefficient of x2 is unity. So, we add and subtract square of half of coefficient of x.
4(x2 – 3x + 5/4) = 4 [x2 – 3x + (3/2)2 – (3/2)2 + 5/4] (Adding and subtracting (3/2)2)
= 4 [(x – 3/2)2 – 12] (Completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= 4 (x – 3/2 + 1) (x – 3/2 – 1)
= 4 (x – 1/2) (x – 5/2) (by taking LCM)
= 4 [(2x-1)/2] [(2x – 5)/2]
= (2x – 5) (2x – 1)
9. y2 – 7y + 12
Solution:
We have,
y2 – 7y + 12
Coefficient of y2 is unity. So, we add and subtract square of half of coefficient of y.
y2 – 7y + 12 = y2 – 7y + (7/2)2 – (7/2)2 + 12 [Adding and subtracting (7/2)2]
= (y – 7/2)2 – (7/2)2 (Completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= (y – (7/2- 1/2)) (y – (7/2 + 1/2))
= (y – 3) (y – 4)
10. z2 – 4z – 12
Solution:
We have,
z2 – 4z – 12
Coefficient of z2 is unity. So, we add and subtract square of half of coefficient of z.
z2 – 4z – 12 = z2 – 4z + 22 – 22 – 12 [Adding and subtracting 22]
= (z – 2)2 – 42 (Completing the square)
By using the formula (a2 – b2) = (a+b) (a-b)
= (z – 2 + 4) (z – 2 – 4)
= (z – 6) (z + 2)
Courtesy : CBSE