RS Aggarwal Solutions Class 8 Chapter-7 Factorization (Ex 7A) Exercise 7.1 (2024)

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RS Aggarwal Chapter 7 Exercise 7.1 Class 8 Solutions

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Factorize :

Question 1.
Solution:
(i) 12x + 15 = 3(4x + 5) {HCF of 12 and 15 = 3)
(ii) 14m – 21 = 7(2m – 3) {HCF of 14 and 21 = 7)
(iii) 9n – 12n² = 3n(3 – 4n) Ans.{HCF of 9 and 12 = 3}

Question 2.
Solution:
(i) 16a² – 24ab = 8a(2a – 3b) { HCF of 16 and 24 = 8)
(ii) 15ab² – 20a²b = 5ab(3b – 4a) {HCF of 15 and 20 5}
(iii) 12x²y³ – 21x³y²= 3x²y²(4y – 7x) Ans. {HCF of 12 and 21 = 3)

Question 3.
Solution:
(i) 24x³ – 36x²y = 12x²(2x – 3y) {HCF of 24 and 36 = 12}
(ii) 10x³ – 15x² = 5x²(2x – 3) {HCF of 10, 15 = 5}
(iii) 36x³y – 60x²y³z = 12x²y (3x – 5y²z) {HCF of 36 and 60 = 12}

Question 4.
Solution:
(i)9x³ – 6x² + 12x
= 3x(3x² – 2x + 4) {HCF of 9, 6, 12 = 3)
(ii)8x² – 72xy + 12x
= 4x(2x – 18y + 3) (HCF of 8, 72, 124)
(iii)18a³b³ – 27a²b³ + 36a³b²
= 9a²b²(2ab – 3b + 4a) Ans.{HCF of 10, 27, 36 = 9)

Question 5.
Solution:
(i) 14x³ + 21x⁴y – 28x²y²
= 7x² (2x + 3x²y – 4y²) {HCF of 14, 21 28 = 7)
(ii) – 5 – 10t + 20t²
= – 5(1 + 2t – 4t²) Ans. {HCF of 5, 10, 20=5)

Question 6.
Solution:
(i) x(x + 3) + 5(x + 3)
=(x + 3)(x + 5)
(ii) 5x(x – 4) – 7(x – 4)
= (x – 4) (5x – 7)
(iii) 2m (1 – n) + 3(1 – n)
= (1 – n) (2m + 3) Ans.

Question 7.
Solution:
6a(a – 2b) + 5b(a – 2b)
= (a – 2b) (6a + 5b) Ans.

Question 8.
Solution:
x³(2a – b) + x²(2a – b)
=x²(2a – b)(x + 1)Ans.

Question 9.
Solution:
9a(3a – 5b) – 12a²(3a – 5b)
= 3a (3a – 5b) (3 – 4a) Ans.

Question 10.
Solution:
(x + 5)² – 4(x + 5)
= (x + 5)(x + 5 – 4)
=(x + 5)(x + 1)Ans.

Question 11.
Solution:
3(a – 2b)² – 5 (a – 2b)
= (a – 2b) {3(a – 2b) – 5}
= (a – 2b) (3a – 6b – 5) Ans.

Question 12.
Solution:
2a + 6b – 3 (a + 3b)²
= 2(a + 3b) – 3(a + 3b)²
= (a + 3b){ 2 – 3 (a + 3b)}
= (a + 3b) (2 – 3a – 9b) Ans.

Question 13.
Solution:
16(2p – 3q)² – 4 (2p – 3q)
= 4(2p – 3q) {4(2p – 3q) – 1}
= 4(2p – 3q) (8p – 12q – 1) Ans.

Question 14.
Solution:
x(a – 3) + y (3 – a)
= x(a – 3) – y(a – 3)
= (a – 3) (x – y) Ans.

Question 15.
Solution:
12(2x – 3y)² – 16(3y – 2x)
= 12(2x – 3y)² + 16(2x – 3y)
= 4(2x – 3y) {3(2x – 3y) + 4}
= 4(2x – 3y) (6x – 9y + 4) Ans.

Question 16.
Solution:
(x + y)(2x + 5) – (x + y)(x + 3)
= (x + y)(2 + 5 – x – 3)
= (x + y)(x + 2) Ans.

Question 17.
Solution:
ar + br + at + bt
= r(a + b) + t(a + b)
= (a + b) (r + t) Ans.

Question 18.
Solution:
x² – ax – bx + ab
= x(x – a) – b(x – a)
= (x – a)(x – b)Ans.

Question 19.
Solution:
ab² – bc² – ab + c²
= ab² – ab – bc² + c²
= ab(b – 1) – c²(b – 1)
= (b – 1) (ab – c²) Ans.

Question 20.
Solution:
x² – xz + xy – yz
= x(x – z) + y(x – z)
= (x – z)(x + y)Ans.

Question 21.
Solution:
6ab – b² + 12ac – 2bc
6ab + 12ac – b² – 2bc
= 6a(b + 2c) – b(b + 2c)
= (b + 2c) (6a – b) Ans.

Question 22.
Solution:
(x – 2y)² + 4x – 8y
= (x – 2y)² + 4(x – 2y)
= (x – 2y)(x – 2y + 4)Ans.

Question 23.
Solution:
y² – xy(1 – x) – x³
y² – xy + x²y – x²
= y(y – x) + x² (y – x)
= (y – x)(y + x²)Ans.

Question 24.
Solution:
(ax + by)² + (bx – ay)²
= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy
= a²x² + b²y² + b²x² + a²y²
= a²x² + b²x² + a²y² + b²y²
= x²(a² + b²) + y²(a² + b²)
= (a² + b²) (x² + y²) Ans.

Question 25.
Solution:
ab² + (a – 1)b – 1
= ab² + ab – b – 1
= ab(b + 1) – 1(b + 1)
= (b + 1) (ab – 1) Ans.

Question 26.
Solution:
x³ – 3x² + x – 3
= x²(x – 3) + 1(x – 3)
(x – 3)(x² + 1)Ans.

Question 27.
Solution:
ab (x² + y²) – xy (a² + b²)
= abx² + aby² + xya² – xyb²
= abx² – xya² – xyb² + aby2
=ax(bx – ay) – by(bx – ay)
= (bx – ay) (ax – by) Ans.

Question 28.
Solution:
x² – x (a + 2b) + 2ab
= x² – xa – 2bx + 2ab
= x(x – a) – 2b(x – a)
= (x – a)(x – 2b)

Hope given RS Aggarwal Class 8 Solutions Chapter 7 Ex 7.1 Factorisation is helpful to complete your CBSE Math homework. If you have any doubts, please comment below.

RS Aggarwal Class 8 Solutions Chapter 7 Ex 7.1- FAQs