RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 8 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 8. Students can also read RBSE Class 8 Maths Important Questions for exam preparation. Students can also go through RBSE Class 8 Maths Notes to understand and remember the concepts easily. Practicing the class 8 maths chapter 6 try these solutions will help students analyse their level of preparation.

RBSE Class 8 Maths Solutions Chapter 14 Factorization Ex 14.2

Question 1.
Factorise the following expressions:
(i) a2 + 8a + 16
Answer:
a2 + 8a + 16
= a2 + 2 × a × 4 + 42
= (a + 4)2
[Using : a2 + 2ab + b2 = (a + b)2]
= (a + 4) (a + 4)

(ii) p2 - 10p + 25
Answer:
p2 - 10p + 25 = p2 - 2 × p × 5 + (5)2
= (P - 5)2
[Using : a2 - 2ab + b2 = (a - b)2]
= (p - 5) (p - 5)

RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

(iii) 25 m2 + 30m + 9
Answer:
25m2 + 30m + 9 = (5m)2 + 2 × 5m × 3 + (3)2
= (5m + 3)2
[Using : a2 + 2ab + b2 = (a + b)2]
= (5m + 3) (5m + 3)

(iv) 49y2 + 84yz + 36z2
Answer:
49y2 + 844yz + 36z2 = (7y)2 + 2 × 7y × 6z + (6z)2
= (7y + 6z)sup>2
[Using : a2 + 2ab + b2 = (a + b)sup>2]
= (7y + 6z)( 7y + 6z)

(v) 4x2 - 8x + 4
Answer:
4x2 - 8x + 4= 4 (x2 - 2x + 1)
= 4(x2 - 2 × x × 1 + 12)
= 4(x - 1)sup>2
[Using : a2 - 2ab + b2 = (a - b)2]
= 4(x - 1)(x - 1)

(vi) 121b2 - 88bc + 16c2
Answer:
121b2 - 88bc +16c2 = (11b)2 - 2 x 11b x 4c + (4c)2
= (11b - 4c)2
[Using : a2 - 2ab + b2 = (a - b)2]
= (11b - 4c)(11b - 4c)

(vii) (l + m)2 - 4lm
Answer:
(l + m)2 - 4lm = l2 + 2lm + m2 - 4lm
= l2 - 11m + m2
= (l - m)2
= (l - m) (l - m)

(viii) a4 + 2a2b2 + b4
Answer:
a4 + 2a2b2 + b4
= (a2)2 + 2 x a2 x b2 + (fb2)2
= (a2 + b2)2
= (a2 + b2)(a2 + b2)

Question 2.
Factorise:
(i) 4p2 - 9q2
Answer:
4p2 - 9q2 = (2p)2 - (3q)2
= (2p + 3q) (2p - 3q)
[Using : a2 - b2 = (a +b) (a - b)]

(ii) 63a2 - 112b2
Answer:
63a2 - 112b2 = 7(9a2 - 16b2)
= 7[(3a)2 - (4b)2]
= 7(3a + 4b)(3a - 4b)
[Using : a2 - b2 = (a + b) (a - b)]

RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

(iii) 49x2 - 36
Answer:
49x2 - 36 = (7x)2 - (6)2
= (7x + 6) (7x - 6)
[Using : a2 - b2 = (a + b) (a - b)]

(iv) 16x5 - 144x3
Answer:
16x5 - 144x3 = 1 6x2(x2 - 9)
= 16x3(x2 - 32)
= (x + 3)(x - 3)
[Using : a2 - b2 = (a + b) (a - b)]

(v) (l + m)2 - (l - m)2
Answer:
(l + m)2 - (l - m)2
[(l + m) + (l - m)] [(l + m) - (l - m)]
[Using : a2 - b2 = (a + b) (a - b)]
= (2l) (2m) = 4lm

(vi) 9x2y2 - 16
Answer:
9x2y2 - 16= (3x)2 - (4)2
= (3xy + 4)(3xy - 4)
[Using : a2 - b2 = (a - b) (a + b)]

(vii) (x2 - 2xy + y2) - z2
Answer:
(x2 - 2xy + y2) - z2
= (x - y)2 - z2
= [(x - y) + z] [(x - y) - z]
= (x - y + z)(x - y - z)

(viii) 25a2 - 4b2 + 28bc - 49c2
Answer:
25a2 - 4b2 + 28bc - 49c2
= 25a2 - (4b2 - 28bc + 49c2)
= 25a2 - [(2b)2 - 2 × 2b × 7c + (7c)2]
= (5a)2 - (2b - 7c)2
= [5a + (2b - 7c)][5a - (2b - 7c)]
= (5a + 2b - 7c)(5a -2b + 7c)

Question 3.
Factorise the expressions:
(i) ax2 + bx
Answer:
ax2 + bx = x(ax + b)

(ii) 7p2 + 21 q2
Answer:
7p2 + 21q2 = 7(p2 + 3q2)

RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

(iii) 2x3 + 2xy2 + 2xz2
Answer:
2x3 + 2xy2 + 2xz2
= 2x(x2 + y2 + z2)

(iv) am2 + bm2 + bn2 + an2
Answer:
am2 + bm2 + bn2 + an2
= (am2 + bm2) + (bn2 + an2)
= (a + b)m2 + (b + a)n2
= (a + b) (m2 + n2)

(v) (lm + l) + m + l
Answer:
(lm + l) + m + 1 = l(m + 1) + 1 (m + 1)
= (m + l) (l + 1)

(vi) y (y + z) + 9(y + z)
Answer:
y(y + z) + 9(y + z)
= (y + z) (y + 9)

(vii) 5y2 - 20y - 8z + 2yz
Answer:
5y2 - 20y - 8z + 2yz
= (5y2 - 20y) + 2yz - 8z
= 5y(y - A) + 2z(y - 4)
= (y - 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2
Answer:
10 ab + 4a + 5b + 2
= (10ab + 5b) + (4a + 2)
= 5b(2a + 1) + 2(2 a + 1)
= (2a + 1) (5b + 2)

(ix) 6xy - 4y + 6 - 9x
Answer:
6xy - 4y + 6 - 9x
= (6xy - 4y) - (9x - 6)
= 2y(3x - 2) - 3(3x - 2)
= (3x - 2)(2y - 3)

RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

Question 4.
Factorise:
(i) a4 - b4 
Answer:
a4 - b4 = (a2)2 - (b2)2
= (a2 + b2) (a2 - b2)
= (a2 + b2) (a + b) (a-b)

(ii) p4 - 81
Answer:
p4 - 81 = (p2)2 - (9)2
= (p2 + 9) (p2 - 9)
= (p2 +9) (p + 3) (p - 3)

(iii) x4 - (y + z)4 
Answer:
x4 - (y + z)4 = (x2)2 - [(y + z)2]2
= [x2 + (y + z)2] [x2 - (y + z)2]
= [x2 + (y + z)2][x + (y + z)] [x - (y + z)]
= [x2 + (y + z)2](x + y + z) (x - y - z)

(iv) x4 - (x - z)4 
Answer:
x4 - (x - z)4 = (x2)2 - [(x - z)2]2
= [x2 + (x - z)2][x2 - (x - z)2]
= [x2 + (x - z)2][x + (x - z)] [x - (x - z)]
= [x2 + (x - z)2] (2x - z) (z)
= z(2x - z) [x2 + (x - z)2]
= z(2x - z) (2x2 - 2xz + z2)

(v) a4 - 2a2b2 + b4
Answer:
a4 - 2a2b2 + b4
= (a2)2 - 2 × a2 × b2 + (b2)2
= (a2 - b2)2
= [(a + b) (a- b)]2
= (a + b) × (a- b)
= (a + b)(a + b)(a - b)(a - b)

Question 5.
Factorise the following expressions:
(i) p2 + 6p + 8
Answer:
p2 + 6p + 8 = (p2 + 6p + 9) - 1
[Using 8 = 9 - 1]
= (p2 + 2 × p × 3 + 32) - 1
= (p + 3)2 - l2 = (p + 3 + 1 )(p + 3 - 1)
= (p + 4)(p + 2)

(ii) q2 - 10q + 21
Answer:
q2 - 10q + 21 = (q2 - 10q + 25) - 4
[Using 21 =25 - 4] = (q2 - 2 × q × 5 + 52) - 4
= (q - 5)2 - 22
= (q - 5 + 2)(q - 5 - 2)
= (q - 3) (q - 7)

RBSE Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

(iii) p2 + 6p - 16
Answer:
p2 + 6p - 16 = (p22 + 6p + 9) - 25
[Using - 16 = 9 - 25]
= (p + 3)2 - 52
= (p + 3 + 5)(p + 3 - 5)
= (p + 8) (p - 2)

Prasanna
Last Updated on May 20, 2022, 2:28 p.m.
Published May 20, 2022