GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1

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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. āˆµ 12x = 2 Ɨ 2 Ɨ 3 Ɨ x = (2 Ɨ 2 Ɨ 3) Ɨ x
and 36 = 2 Ɨ 2 Ɨ 3 Ɨ 3 = (2 Ɨ 2 Ɨ 3) Ɨ 3
āˆ“ the common factor = 2 Ɨ 2 Ɨ 3 = 12

2. āˆµ 2y = 2 Ɨ y = (2 Ɨ y)
and 22y = 2 Ɨ 11 Ɨ y
= (2 Ɨ y) Ɨ 11
āˆ“ the common factor = 2 Ɨ y = 2y

3. āˆµ 14pq = 2 Ɨ 7 Ɨ p Ɨ q = (2 Ɨ 7 Ɨ p Ɨ q)
and 28 p2q2 = 2 Ɨ 2 Ɨ 1 Ɨ p Ɨ p Ɨ q Ɨ q
= (2 Ɨ 7 Ɨ p Ɨ q) Ɨ 2 Ɨ p Ɨ q
āˆ“ the common factor = (2 Ɨ 7 Ɨ p Ɨ q) =14pq

4. āˆµ 2x = 1 Ɨ 2 Ɨ x = 1 Ɨ 2 Ɨ x
3x2 = 1 Ɨ 3 Ɨ x Ɨ x = 1 Ɨ 3 Ɨ x and
4 = 1 Ɨ 2 Ɨ 2 = 1 Ɨ 2 Ɨ 2
āˆ“ the common factor = 1
Note: 1 is a factor of every term.

5. āˆµ 6 abc = 2 Ɨ 3 Ɨ a Ɨ b Ɨ c
= (2 Ɨ 3 Ɨ a Ɨ b) Ɨ c
24 ab2 = 2 Ɨ 2 Ɨ 2 Ɨ 3 Ɨ a Ɨ b Ɨ b
= (2 Ɨ 3 Ɨ a Ɨ b) Ɨ 2 Ɨ 2 Ɨ b
12 a2b = 2 Ɨ 2 Ɨ 3 Ɨ a Ɨ a Ɨ b
= (2 Ɨ 3 Ɨ a Ɨ b) Ɨ 2 Ɨ a
āˆ“ the common factor

6. āˆµ 10pq = 2 Ɨ 5 Ɨ p Ɨ q
= (2 Ɨ 5) Ɨ p Ɨ q
20qr = 2 Ɨ 2 Ɨ 5 Ɨ q Ɨ r
= (2 Ɨ 5) Ɨ 2 Ɨ q Ɨ r
30rp = 2 Ɨ 3 Ɨ 5 Ɨ r Ɨ p
āˆ“ the common factor = 2 Ɨ 5 = 10

7. āˆµ 3x2y3 = 3 Ɨ x Ɨ x Ɨ y Ɨ y Ɨ y
= (x Ɨ x Ɨ y Ɨ y) Ɨ 3 Ɨ y
10x3y2 = 2 Ɨ 5 Ɨ x Ɨ x Ɨ x Ɨ y Ɨ y
= (x Ɨ x Ɨ y Ɨ y) Ɨ 2 Ɨ 5 Ɨ x
6x2y2z = 2 Ɨ 3 Ɨ x Ɨ x Ɨ y Ɨ y Ɨ z
= (x Ɨ x Ɨ y Ɨ y) Ɨ 2 Ɨ 3 Ɨ z
āˆ“ the common factor = (x Ɨ x Ɨ y Ɨ y) = x2y2

GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1

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. āˆµ 7x = 7 Ɨ x = (7) Ɨ x
42 = 2 Ɨ 7 Ɨ 3 = (7) Ɨ 2 Ɨ 3
āˆ“ 7x – 42 = 7[(x)] – (2 Ɨ 3)] = 7[x – 6]

2. āˆµ 6p = 2 Ɨ 3 Ɨ p
= (2) Ɨ (3) Ɨ p = (2 Ɨ 3) Ɨ p
12q = 2 Ɨ 2 Ɨ 3 Ɨ q
= (2) Ɨ 2 Ɨ (3) Ɨ q
= (2 Ɨ 3) Ɨ 2 Ɨ q
āˆ“ 6p – 12q = (2 Ɨ 3) [(p) – (2 Ɨ q)]
= 6[p – 2q]

3. āˆµ 7a2 = 7 Ɨ a Ɨ a = (7 Ɨ a) Ɨ a
14a = 2 Ɨ 7 Ɨ a = (7 Ɨ a) Ɨ 2
āˆ“ 7a2 – 14a = (7 Ɨ a)[a – 2] = 7a(a – 2)

4. āˆµ -16z = (-1) Ɨ 2 Ɨ 2 Ɨ 2 Ɨ 2 Ɨ z
= (2 Ɨ 2 Ɨ z) Ɨ (-1) Ɨ 2 Ɨ 2
20z3 = 2 Ɨ 2 Ɨ 5 Ɨ z Ɨ z Ɨ z
= (2 Ɨ 2 Ɨ z) Ɨ 5 Ɨ z Ɨ z
āˆ“ -16z + 20z3 = (2 Ɨ 2 Ɨ z) [(-1) Ɨ 4 + 5 Ɨ z Ɨ z] = 4z[-4 + 5z2]

5. āˆµ 20l2m = 2 Ɨ 2 Ɨ 5 Ɨ l Ɨ l Ɨ m
= (2 Ɨ 5 Ɨ l Ɨ m) Ɨ 2 Ɨ l
30alm = 2 Ɨ 3 Ɨ 5 Ɨ a Ɨ l Ɨ m
= (2 Ɨ 5 Ɨ l Ɨ m) Ɨ 3 Ɨ a
āˆ“ 2ol2m + 30alm = (2 Ɨ 5 Ɨ l Ɨ m) [2 Ɨ l + 3 Ɨ a] = 10lm [2l + 3a]

6. āˆµ 5x2y = 5 Ɨ x Ɨ x Ɨ y
= (5 Ɨ x Ɨ y) [x]
15x2y = 5 Ɨ 3 Ɨ x Ɨ y Ɨ y
= (5 Ɨ x Ɨ y) [3 Ɨ y]
āˆ“ 5x2y – 15xy2 = (5 Ɨ x Ɨ y)[x – 3 Ɨ y]
= 5xy(x – 3y)

7. āˆµ 10a2 = 2 Ɨ 5 Ɨ a Ɨ a
= (5) [2 Ɨ a Ɨ a]
15b2 = 3 Ɨ 5 Ɨ b Ɨ b
= (5) [3 Ɨ b Ɨ b]
20c2 = 2 Ɨ 2 Ɨ 2 Ɨ 5 Ɨ c Ɨ c
= (5) [2 Ɨ 2 Ɨ c Ɨ c]
āˆ“ 10a2 – 15b2 + 20c2 = (5)[2 Ɨ a Ɨ a – 3 Ɨ b Ɨ b + 2 Ɨ 2 Ɨ c Ɨ c]
= 5[2a2 – 3b2 + 4c2]

8. āˆµ -4a2 = (-1) Ɨ 2 Ɨ 2 Ɨ a Ɨ a
= (2 Ɨ 2 Ɨ a) [(-1) Ɨ a]
4ab = 2 Ɨ 2 Ɨ a Ɨ b
= (2 Ɨ 2 Ɨ a)[b]
-4ca = (-1) Ɨ 2 Ɨ 2 Ɨ c Ɨ a
= (2 Ɨ 2 Ɨ a)[-1) Ɨ c]
āˆ“ -4a2 + 4ab – 4ca = (2 Ɨ 2 Ɨ a) [(-1)] Ɨ a + b – c]
= 4a[-a + b – c]

9. āˆµ x2yz = x Ɨ x Ɨ y Ɨ z = (xyz)[x]
xy2z = x Ɨ y Ɨ y Ɨ z = (xyz)[y]
xyz2 = x Ɨ y Ɨ z Ɨ z = (xyz)[z]
āˆ“ x2yz + xy2z + xyz2 = (xyz)[x + y + z]
= xyz(x + y + z)

10. āˆµ ax2y = a Ɨ x Ɨ x Ɨ y = (x Ɨ y)[a Ɨ x]
bxy2 = b Ɨ x Ɨ y Ɨ y
= (x Ɨ y)[a Ɨ x + b Ɨ y + c Ɨ z]
= xy(ax + by + cz)

GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.1

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)

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