# GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1

Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1

Question 1.
Find the common factors of the given terms?

1. 12x, 36
2. 2y, 22xy
3. 4pq, 28p2q2
4. 2x, 3x2, 4
5. 6abc, 24ab2, 12a2b
6. 6x3, -4x2, 32x
7. 10pq, 20qr, 30rp
8. 3x2y3, 10x3y2, 6x2y2z

Solution:
1. āµ 12x = 2 Ć 2 Ć 3 Ć x = (2 Ć 2 Ć 3) Ć x
and 36 = 2 Ć 2 Ć 3 Ć 3 = (2 Ć 2 Ć 3) Ć 3
ā“ the common factor = 2 Ć 2 Ć 3 = 12

2. āµ 2y = 2 Ć y = (2 Ć y)
and 22y = 2 Ć 11 Ć y
= (2 Ć y) Ć 11
ā“ the common factor = 2 Ć y = 2y

3. āµ 14pq = 2 Ć 7 Ć p Ć q = (2 Ć 7 Ć p Ć q)
and 28 p2q2 = 2 Ć 2 Ć 1 Ć p Ć p Ć q Ć q
= (2 Ć 7 Ć p Ć q) Ć 2 Ć p Ć q
ā“ the common factor = (2 Ć 7 Ć p Ć q) =14pq

4. āµ 2x = 1 Ć 2 Ć x = 1 Ć 2 Ć x
3x2 = 1 Ć 3 Ć x Ć x = 1 Ć 3 Ć x and
4 = 1 Ć 2 Ć 2 = 1 Ć 2 Ć 2
ā“ the common factor = 1
Note: 1 is a factor of every term.

5. āµ 6 abc = 2 Ć 3 Ć a Ć b Ć c
= (2 Ć 3 Ć a Ć b) Ć c
24 ab2 = 2 Ć 2 Ć 2 Ć 3 Ć a Ć b Ć b
= (2 Ć 3 Ć a Ć b) Ć 2 Ć 2 Ć b
12 a2b = 2 Ć 2 Ć 3 Ć a Ć a Ć b
= (2 Ć 3 Ć a Ć b) Ć 2 Ć a
ā“ the common factor

6. āµ 10pq = 2 Ć 5 Ć p Ć q
= (2 Ć 5) Ć p Ć q
20qr = 2 Ć 2 Ć 5 Ć q Ć r
= (2 Ć 5) Ć 2 Ć q Ć r
30rp = 2 Ć 3 Ć 5 Ć r Ć p
ā“ the common factor = 2 Ć 5 = 10

7. āµ 3x2y3 = 3 Ć x Ć x Ć y Ć y Ć y
= (x Ć x Ć y Ć y) Ć 3 Ć y
10x3y2 = 2 Ć 5 Ć x Ć x Ć x Ć y Ć y
= (x Ć x Ć y Ć y) Ć 2 Ć 5 Ć x
6x2y2z = 2 Ć 3 Ć x Ć x Ć y Ć y Ć z
= (x Ć x Ć y Ć y) Ć 2 Ć 3 Ć z
ā“ the common factor = (x Ć x Ć y Ć y) = x2y2

Question 2.
Factorise the following expressions:

1. 7x – 42
2. 6p – 12q
3. 7a2 + 14a
4. -16z + 20z3
5. 20l2m + 30alm
6. 5x2y – 15xy2
7. 10 a2 – 15 b2 + 20 c2
8. -4a2 + 4ab – 4ca
9. x2yz + xy2z + xyz2
10. ax2y + bxy2 + cxyz

Solution:
1. āµ 7x = 7 Ć x = (7) Ć x
42 = 2 Ć 7 Ć 3 = (7) Ć 2 Ć 3
ā“ 7x – 42 = 7[(x)] – (2 Ć 3)] = 7[x – 6]

2. āµ 6p = 2 Ć 3 Ć p
= (2) Ć (3) Ć p = (2 Ć 3) Ć p
12q = 2 Ć 2 Ć 3 Ć q
= (2) Ć 2 Ć (3) Ć q
= (2 Ć 3) Ć 2 Ć q
ā“ 6p – 12q = (2 Ć 3) [(p) – (2 Ć q)]
= 6[p – 2q]

3. āµ 7a2 = 7 Ć a Ć a = (7 Ć a) Ć a
14a = 2 Ć 7 Ć a = (7 Ć a) Ć 2
ā“ 7a2 – 14a = (7 Ć a)[a – 2] = 7a(a – 2)

4. āµ -16z = (-1) Ć 2 Ć 2 Ć 2 Ć 2 Ć z
= (2 Ć 2 Ć z) Ć (-1) Ć 2 Ć 2
20z3 = 2 Ć 2 Ć 5 Ć z Ć z Ć z
= (2 Ć 2 Ć z) Ć 5 Ć z Ć z
ā“ -16z + 20z3 = (2 Ć 2 Ć z) [(-1) Ć 4 + 5 Ć z Ć z] = 4z[-4 + 5z2]

5. āµ 20l2m = 2 Ć 2 Ć 5 Ć l Ć l Ć m
= (2 Ć 5 Ć l Ć m) Ć 2 Ć l
30alm = 2 Ć 3 Ć 5 Ć a Ć l Ć m
= (2 Ć 5 Ć l Ć m) Ć 3 Ć a
ā“ 2ol2m + 30alm = (2 Ć 5 Ć l Ć m) [2 Ć l + 3 Ć a] = 10lm [2l + 3a]

6. āµ 5x2y = 5 Ć x Ć x Ć y
= (5 Ć x Ć y) [x]
15x2y = 5 Ć 3 Ć x Ć y Ć y
= (5 Ć x Ć y) [3 Ć y]
ā“ 5x2y – 15xy2 = (5 Ć x Ć y)[x – 3 Ć y]
= 5xy(x – 3y)

7. āµ 10a2 = 2 Ć 5 Ć a Ć a
= (5) [2 Ć a Ć a]
15b2 = 3 Ć 5 Ć b Ć b
= (5) [3 Ć b Ć b]
20c2 = 2 Ć 2 Ć 2 Ć 5 Ć c Ć c
= (5) [2 Ć 2 Ć c Ć c]
ā“ 10a2 – 15b2 + 20c2 = (5)[2 Ć a Ć a – 3 Ć b Ć b + 2 Ć 2 Ć c Ć c]
= 5[2a2 – 3b2 + 4c2]

8. āµ -4a2 = (-1) Ć 2 Ć 2 Ć a Ć a
= (2 Ć 2 Ć a) [(-1) Ć a]
4ab = 2 Ć 2 Ć a Ć b
= (2 Ć 2 Ć a)[b]
-4ca = (-1) Ć 2 Ć 2 Ć c Ć a
= (2 Ć 2 Ć a)[-1) Ć c]
ā“ -4a2 + 4ab – 4ca = (2 Ć 2 Ć a) [(-1)] Ć a + b – c]
= 4a[-a + b – c]

9. āµ x2yz = x Ć x Ć y Ć z = (xyz)[x]
xy2z = x Ć y Ć y Ć z = (xyz)[y]
xyz2 = x Ć y Ć z Ć z = (xyz)[z]
ā“ x2yz + xy2z + xyz2 = (xyz)[x + y + z]
= xyz(x + y + z)

10. āµ ax2y = a Ć x Ć x Ć y = (x Ć y)[a Ć x]
bxy2 = b Ć x Ć y Ć y
= (x Ć y)[a Ć x + b Ć y + c Ć z]
= xy(ax + by + cz)

Question 3.
Factorise:

1. x2 + xy + 8x + 8y
2. 15xy – 6x + 5y – 2
3. ax + bx – ay – by
4. 15pq + 15 + 9q + 25p
5. z – 7 + 7xy – xyz

Solution:
1. x2 + xy + 8x + 8y
= x[x + y] + 8[x + y]
= (x + y)(x + 8)

2. 15xy – 6x + 5y – 2
= 3x[5y – 2] + 1[5y – 2]
= (5y – 2)(3x + 1)

3. ax + bx – ay – by
= x[a + b] + (-y)[a + b]
= (x – y)[a + b]

4. Regrouping the terms, we have
15pq + 15 + 9q + 25p
= 15pq + 9q + 25p + 15
= 3q[5p + 3] + 5[5p + 3]
= (5p + 3)[3q + 5]

5. Regrouping the terms, we have
z – 7 + 7xy – xyz = z – 7 – xyz + 7xy
= 1[z – 7] – xy[z – 7] = (z – 7)(1 – xy)