GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2

March 1, 2023 / By / Class 8

Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 1.
Factorise the following expressions.

  1. a2 + 8a + 16
  2. p2 – 10p + 25
  3. 25m2 + 30m + 9
  4. 49y2 + 84yz + 36z2
  5. 4x2 – 8x + 4
  6. 121b2 – 88bc + 16c2
  7. (1+ m)2 – 41m
  8. a4 + 2a2b2 + b4 [Hint: Expand (1 + m)2 first].

Solution:
1. We have a2 + 8a + 16
= (a)2 + 2(a)(4) + (4)2
= (a + 4)2 = (a + 4)(a + 4)
āˆ“ a2 + 8a + 16 = (a + 4)2 = (a + 4)(a + 4)

2. We have p2 – 10p + 25
= p2 – 2(p)(5) + (5)2
= (p – 5)2 = (p – 5)(p – 5)

3. We have 25m2 + 30m + 9
= (5m)2 + 2(5m)(3) + (3)2
= (5m + 3)2 = (5m + 3)(5m + 3)
āˆ“ 25m2 + 30m + 9
= (5m + 3)2 = (5m + 3)(5m + 3)

4. We have 49y2 + 84yz + 36z
= (7y)2 + 2(7y)(6z) + (6z)2
= (7y + 6z)2
= (7y + 6z)(7y + 6z)
āˆ“ 49y2 + 84yx + 36z2
= (7y + 6z)2
= (7y + 6z)(7y + 6z)

5. We have 4x2 – 8x + 4
= (2x)2 – 2(2x)(2) + (2)2
= (2x – 2)2 = (2x – 2)(2x – 2)
= 2(x – 1)2(x – 1)
= 4(x – 1)(x – 4)
āˆ“ 4x2 – 8x + 4 = 4(x – 1)(x – 1)
= 4(x – 1)2

6. We have 121b2 – 88bc + 16c2
= (11b)2 – 2(11b)(4c) + (4c)2
= (11b – 4c)2 = (11b – 4c)(11b – 4c)
āˆ“ 121b2 – 88bc + 16c2

7. We have (l + m)2 – 4lm
= (l2 + 2lm + m2) – 4lm
[Collecting the like terms 2lm and -4lm]
= l2 + (2lm – 4lm) + m2
= l2 + 2lm + m2
= (l)2 – 2(l)(m) + (m)2
= (l – m)2 = (l – m)(l – m)
āˆ“ (l + m)2 – 4lm = (l – m)2
= (l – m)(l – m)

8. We have a4 + 2a2b2 + b4
= (a2)2 + 2(a2)(b2) + (b2)2
= (a2 + b2)2 = (a2 + b2)(a2 + b2)
āˆ“ a4 + 2a2b2 + b2
= (a2 + b2)2 = (a2 + b2)(a2 + b2)

GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 2.
Factorise:

  1. 4p2 – 9q2
  2. 63a2 – 112b2
  3. 49x2 – 36
  4. 16x5 – 144x3
  5. (l + m)2 – (l – m)2
  6. 9x2y2 – 16
  7. (x2 – 2xy + y2) – z2
  8. 25a2 – 4b2 + 28bc – 49c2

Solution:
1. āˆµ 4p2 – 9q2 = (2p)2 – (3q)2
= (2p – 3q)(2p + 3q)
[Using a2 – b2 (a + b)(a – b)
āˆ“ 4p2 – 9q2 = (2p – 3q)(2p + 3q)

2. We have
63a2 – 112b2 = 7 Ɨ 9a2 – 7 Ɨ 16b2
= 7(9a2 – 16b2)
āˆµ 9a2 – 16b2 = (3a)2 – (4b)2
= (3a + 4b)(3a – 4b)
[Using a2 – b2 = (a + b)(a – b)]
āˆ“ 63a2 – 112b2 = 7[(3a) + 4b)(3a – 4b)]

3. āˆµ 49x2 – 36 = (7x)2 – (6)2
= (7x – 6)(7x + 6)
[(Using a2 – b2 = (a – b)(a + b)]
āˆ“ 49x2 – 36 = (7x – 6)(7x + 6)

4. We have 16x5 = 16x3 Ɨ x2 and 144x3
= 16x3 Ɨ 9
āˆ“ 16x5 – 144x3 = (16x3) Ɨ x2 – (16x3) Ɨ 9
= 16x3 Ɨ 9
= 16x3[x2 – 9]
= 16x3[(x)2 – (3)2]
= 16x3[(x + 3)(x – 3)]
[(Using a2 – b2 = (a – b)(a + b)]
āˆ“ 16x5 – 144x3 = 16x3(x + 3)(x – 3)

5. Using the identiy
a2 – b2 = (a + b)(a – b)
we have (l + m)2 – (l – m)2
= [(l + m) + (l – m)][(l + m) – (l – m)]
= [l + m + l – m][l + m – l + m]
= (2l)(2m) = 2 Ɨ 2(l Ɨ m) = 4lm

6. We have 9x2y2 – 16 (3xy)2 – (4)2 = (3xy + 4)(3xy – 4)
[Using a2 – b2 = (a + b)(a – b)]
āˆ“ 9x2y2 – 16 = (3xy + 4)(3xy – 4)

7. We have x2 – 2xy + y2 = (x – y)2
āˆ“ (x2 – 2xy + y2) – z2 = (x – y)2 – (z)2
= [(x – y) + z][(x – y) – z]
[Using a2 – b2 = (a + b)(a – b)]
= (x – y + z)(x – y – z)
āˆ“ (x2 – 2xy + y2) – z2 = (x – y + z)(x – y – z)

8. We have -4b2 + 28bc – 49c2
= (-1)[4b2 – 28bc + 49c2]
= -1[(2b)2 – 2(2b)(7c) + (7c)2]
= -[2b – 7c]2
āˆ“ 25a2 – 4b2 + 28bc – 49c2 = 25a2 – (2b – 7c)2
Now, using a2 – b2 = (a + b)(a – b),
we have [5a]2 – [2b – 7c]2
= (5a + 2b – 7c)(5a – 2b + 7c)
āˆ“ 25a2 – 4a2 + 28bc – 49c2
= (5a + 2b – 7c)(5a – 2b + 7c)

GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 3.
Factorise the expressions?

  1. ax2 + bx
  2. 7p2 + 21q2
  3. 2x3 + 2xy2 + 2xz2
  4. am2 + bm2 + bn2 + an2
  5. (lm + l) + m + l
  6. y(y + z) + 9(y + z)
  7. 5y2 – 20y – 8z + 2yz
  8. 10ab + 4a + 5b + 2
  9. 6xy – 4y + 6 – 9x

Solution:
1. āˆµ ax2 = a Ɨ x Ɨ x = (x)[ax]
bx = b Ɨ x = (x)[b]
āˆ“ ax2 + bx = x[ax + b]

2. We have 7p2 + 21q2
= 7 Ɨ p Ɨ p + 7 Ɨ 3 Ɨ q Ɨ q
= 7[p Ɨ p + 3 Ɨ q Ɨ q]
= 7[p2 + 3q2]

3. Taking out 2x as common from each term, we have
2x3 + 2xy2 + 2xz2= 2x[x2 + y2 + z2]

4. We can take out m2 as common from the first two terms and n2 as common from the last two terms,
āˆ“ am2 + bm2 + bn2 + an2
= m2(a + b) + n2(b + a)
= m2(a + b) + n2(a + b)
= (a + b)[m2 + n2]

5. We have lm + l = l(m + 1)
āˆ“ (lm + l) + (m + 1)
= l(m + 1) + (m + 1)
= (m + 1)[l + 1]

6. We have (y + z) as a common factor to both terms.
āˆ“ y(y + z) + 9(y + z) = (y + z)(y + 9)

7. We have 5y2 – 20y = 5y(y – 4)
and -8z + 2yz = 2z(-4 + y)
= 2z(y – 4)
āˆ“ 5y2 – 20y – 8z + 2yz
= 5y(y – 4) + 2z(y – 4)
= (y – 4)[5y + 2z]

8. We have, 10ab + 4a = 2a(5b + 2)
and (5b + 2) = 1(5b + 2)
āˆ“ 10ab + 4a + 5b + 2
= 2a(5b + 2) + 1(5b + 2)
= (5b + 2)[2a + 1]

9. Regrouping the terms, we get
6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6
= 2y[3x – 2] – 3[3x – 2]
= (3x – 2)[2y – 3]

GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 4.
Factorise:

  1. a4 – b4
  2. p4 – 81
  3. x4 – (y + z)4
  4. x4 – (x – z)4
  5. a4 – 2a2b2 + b4

Solution:
1. Using a2 – b2 = (a – b)(a + b),
we have a4 – b4 = (a2)2 – (b2)2
= (a2 + b2)(a2 – b2)
= (a2 + b2)[(a + b)(a – b)]
= (a2 + b2)(a + b)(a – b)

2. We have p4 – 81 = (p2)2 – (9)2
Now using a2 – b2 = (a + b)(a – b),
we have (p2)2 – (9)2 = (p2 + 9) (p2 – 9)
We can factorise p2 – 9 further as
p2 – 9 = (p)2 – (3)2
= (p + 3)(p – 3)
āˆ“ p4 – 81 = (p + 3)(p – 3)(p2 + 9)

3. āˆµ x4 – (y + z)4 = [x2]2 – [(y + z)2]2
= [(x)2 + (y + z)2][(x2) – (y + z)2]
Using a2 – b2 = (a + b)(a – b)]
We can factorise [x2 – (y + z)2] further as
(x)2 – (y + z)2
= [(x) + (y + z)][(x) – (y + z)]
= (x + y + z)(x – y + z)
āˆ“ x4 – (y + z)4
= (x + y + z)(x – y – z)[x2 + (y + z)2]

4. We have x4 – (x – z)4 = [x2]2 – [(x – z)2]2
= [x2 + (x – z)2][x2 – (x – z)2]
Now, factorising x2 – (x – z)2 further,
we have
x2 – (x – z)2 = [x + (x – z)[x – (x – z)]
= (x + x – z)(x – x + z)
= (2x – z)(z)
āˆ“ x4 – (y – z)4 = z(2x – 2)[x2 + (x – 2)2]
= z(2x – 2)[x2 + (x2 – 2xz + z2)]
= z(2x – z)[x2 + x2 – 2xz + z2]
= z(2x – z)(2x2 – 2xz + z2)

5. āˆµ a4 – 2a2b2 + b4
= (a2)2 – 2(a2)(b2) + (b2)2
= (a2 – b2)2 = [(a2 – b2)(a2 + b2)]
= [(a – b)(a + b)(a2 + b2)]
āˆ“ a4 – 2a2b2 + b4 = (a – b)(a + b)(a2 + b2)

GSEB Solutions Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 5.
Factorise the following expressions.

  1. p2 + 6p + 8
  2. q2 – 10q + 21
  3. p2 + 6p – 16

Solution:
1. We have p2 + 6p + 8 = p2 + 6p + 9 – 1 [āˆµ 8 = 9 – 1]
= [(p)2 + 2(p)(3) + 32] – 1 = (p + 3)2 – 12
[Using a2 + 2ab + b2 = (a + b)2]
= [(p + 3) + 1][(p + 3) – 1]
= (p + 4)(p + 2)
āˆ“ p2 + 6p + 8 = (p + 4)(p + 2)

2. We have q2 – 10q + 21
= q2 – 10q + 25 – 4 [āˆµ 21 = 25 – 4]
= [(q)2 – 2(q)(5) + (5)2] – (2)2
= [q – 5]2 – [2]2
= [(q – 5) + 2][(q – 5) – 2]
[Using a2 – b2 = (a + b)(a – b)] = (q – 3)(q – 7)

3. We have p2 + 6p – 16 = p2 + 6p + 9 – 25 [āˆµ -16 = 9 – 25]
= [(p)2 + 2(p)(3) + (3)2] – (5)2
= [p + 3]2 – [5]2
[Using a2 + 2ab + b2 = (a + b)2]
= [(p + 3) + 5][(p + 3) – 5]
[Using a2 – b2 = (a + b)(a – b)]
= (p + 8)(p – 2)

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