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)