# GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1

Gujarat Board 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. ∵ 12x = 2 × 2 × 3 × x = (2 × 2 × 3) × x
and 36 = 2 × 2 × 3 × 3 = (2 × 2 × 3) × 3
∴ the common factor = 2 × 2 × 3 = 12

2. ∵ 2y = 2 × y = (2 × y)
and 22y = 2 × 11 × y
= (2 × y) × 11
∴ the common factor = 2 × y = 2y

3. ∵ 14pq = 2 × 7 × p × q = (2 × 7 × p × q)
and 28 p2q2 = 2 × 2 × 1 × p × p × q × q
= (2 × 7 × p × q) × 2 × p × q
∴ the common factor = (2 × 7 × p × q) =14pq

4. ∵ 2x = 1 × 2 × x = 1 × 2 × x
3x2 = 1 × 3 × x × x = 1 × 3 × x and
4 = 1 × 2 × 2 = 1 × 2 × 2
∴ the common factor = 1
Note: 1 is a factor of every term.

5. ∵ 6 abc = 2 × 3 × a × b × c
= (2 × 3 × a × b) × c
24 ab2 = 2 × 2 × 2 × 3 × a × b × b
= (2 × 3 × a × b) × 2 × 2 × b
12 a2b = 2 × 2 × 3 × a × a × b
= (2 × 3 × a × b) × 2 × a
∴ the common factor

6. ∵ 10pq = 2 × 5 × p × q
= (2 × 5) × p × q
20qr = 2 × 2 × 5 × q × r
= (2 × 5) × 2 × q × r
30rp = 2 × 3 × 5 × r × p
∴ the common factor = 2 × 5 = 10

7. ∵ 3x2y3 = 3 × x × x × y × y × y
= (x × x × y × y) × 3 × y
10x3y2 = 2 × 5 × x × x × x × y × y
= (x × x × y × y) × 2 × 5 × x
6x2y2z = 2 × 3 × x × x × y × y × z
= (x × x × y × y) × 2 × 3 × z
∴ the common factor = (x × x × y × 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. ∵ 7x = 7 × x = (7) × x
42 = 2 × 7 × 3 = (7) × 2 × 3
∴ 7x – 42 = 7[(x)] – (2 × 3)] = 7[x – 6]

2. ∵ 6p = 2 × 3 × p
= (2) × (3) × p = (2 × 3) × p
12q = 2 × 2 × 3 × q
= (2) × 2 × (3) × q
= (2 × 3) × 2 × q
∴ 6p – 12q = (2 × 3) [(p) – (2 × q)]
= 6[p – 2q]

3. ∵ 7a2 = 7 × a × a = (7 × a) × a
14a = 2 × 7 × a = (7 × a) × 2
∴ 7a2 – 14a = (7 × a)[a – 2] = 7a(a – 2)

4. ∵ -16z = (-1) × 2 × 2 × 2 × 2 × z
= (2 × 2 × z) × (-1) × 2 × 2
20z3 = 2 × 2 × 5 × z × z × z
= (2 × 2 × z) × 5 × z × z
∴ -16z + 20z3 = (2 × 2 × z) [(-1) × 4 + 5 × z × z] = 4z[-4 + 5z2]

5. ∵ 20l2m = 2 × 2 × 5 × l × l × m
= (2 × 5 × l × m) × 2 × l
30alm = 2 × 3 × 5 × a × l × m
= (2 × 5 × l × m) × 3 × a
∴ 2ol2m + 30alm = (2 × 5 × l × m) [2 × l + 3 × a] = 10lm [2l + 3a]

6. ∵ 5x2y = 5 × x × x × y
= (5 × x × y) [x]
15x2y = 5 × 3 × x × y × y
= (5 × x × y) [3 × y]
∴ 5x2y – 15xy2 = (5 × x × y)[x – 3 × y]
= 5xy(x – 3y)

7. ∵ 10a2 = 2 × 5 × a × a
= (5) [2 × a × a]
15b2 = 3 × 5 × b × b
= (5) [3 × b × b]
20c2 = 2 × 2 × 2 × 5 × c × c
= (5) [2 × 2 × c × c]
∴ 10a2 – 15b2 + 20c2 = (5)[2 × a × a – 3 × b × b + 2 × 2 × c × c]
= 5[2a2 – 3b2 + 4c2]

8. ∵ -4a2 = (-1) × 2 × 2 × a × a
= (2 × 2 × a) [(-1) × a]
4ab = 2 × 2 × a × b
= (2 × 2 × a)[b]
-4ca = (-1) × 2 × 2 × c × a
= (2 × 2 × a)[-1) × c]
∴ -4a2 + 4ab – 4ca = (2 × 2 × a) [(-1)] × a + b – c]
= 4a[-a + b – c]

9. ∵ x2yz = x × x × y × z = (xyz)[x]
xy2z = x × y × y × z = (xyz)[y]
xyz2 = x × y × z × z = (xyz)[z]
∴ x2yz + xy2z + xyz2 = (xyz)[x + y + z]
= xyz(x + y + z)

10. ∵ ax2y = a × x × x × y = (x × y)[a × x]
bxy2 = b × x × y × y
= (x × y)[a × x + b × y + c × 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)