RS Aggarwal Class 8 Math Seventh Chapter Factorisation Exercise 7A Solution

RS Aggarwal Class 8 Math Seventh Chapter Factorisation Exercise 7A Solution

EXERCISE 7A

Factorise:

(1) (i) 12x + 15

= [(3×4) x] + (3 × 5)

= 3(4x + 5)

(ii) 14m – 21

= 7 (2m – 3)

(iii) 9n – 12n2

= 3n (3 – 4n)

(2) (i) 16a2 – 24ab

= 8a (2a – 3b)

(ii) 15ab2 – 20a2b

= 5ab (3b – 4a)

(iii) 12x2y3 – 21x3y2

= 3x2y2 (4y – 7x)

(3) (i) 24x3 – 36x2y

= 12x2 (2x – 3y)

(ii) 10x3 – 15x2

= 5x2 (2x – 3)

(iii) 36x3y – 60x2y3z

= 12x2y (3x – 5y2z)

(4) (i) 9x3 – 6x2 + 12x

= 3x (3x2 – 2x + 4)

(ii) 8x2 – 72xy + 12x

= 4x (2x – 18y + 3)

(iii) 18a3b3 – 27a2b3 + 36a3b2

= 9a2b2 (2ab – 3b + 4a)

(5) (i) 14x3 + 21x4y – 28x2y2

= 7x2 (2x + 3x2y – 4y2)

(ii) – 5 – 10t + 20t2

= – 5 (1 + 2y – 4t2)

(6) (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)

(7) 6a(a – 2b) + 5b(a – 2b)

= (a – 2b) (6a + 5b)

(8) x3(2a – b) + x2 (2a – b)

= (2a – b) (x3 + x2)

= x2 (2a – b) (x + 1)

(9) 9a(3a – 5b) – 12a2(3a – 5b)

= (3a – 5b) [3a(3 – 4a)]

= 3a (3a – 5b) (3 – 4a)

(10) (x + 5)2 – 4(x + 5)

= (x + 5) [(x + 5) – 4]

= (x + 5) (x + 1)

(11) 3(a – 2b)2 – 5(a – 2b)

= (a – 2b) (3a – 6b – 5)

(12) 2a + 6b – 3(a + 3b)2

= 2(a + 3b) – 3(a + 3b)2

= (a + 3b) (2 – 3a – 9b)

(13) 16(2p – 3q)2 – 4(2p – 3q)

= (2p – 3q) (32p – 48q – 4)

= 4 (2p – 3q) (8p – 12q – 1)

(14) x(a – 3) + y(3 – a)

= x(a – 3) – y(a – 3)

= (a – 3) (x – y)

(15) 12(2x – 3y)2 – 16(3y – 2x)

= 12(2x – 3y)2 + 16(2x – 3y)

= (2x – 3y) (24x – 36y + 16)

= 4 (2x – 3y) (6x – 9y + 4)

(16) (x + y) (2x + 5) – (x + y) (x + 3)

= (x + y) [(2x + 5) – (x + 3)]

= (x + y) (2x + 5 – x – 3)

= (x + y) (x + 2)

(17) ar + br + at + bt

= r (a + b) + t (a + b)

= (a + b) (r + t)

(18) x2 – axbx + ab

= x (x – a) – b (x – a)

= (x – a) (x – b)

(19) ab2 – bc2ab + c2

= b (ab – c2) – 1(ab – c2)

= (ab – c2) (b – 1)

(20) x2 – xz + xy – yz

= x (x – z) + y (x – z)

= (x – z) (x + y)

(21) 6ab – b2 + 12ac – 2bc

= b(6a – b) + 2c (6a – b)

= (6a – b) (b + 2c)

(22) (x – 2y)2 + 4x – 8y

= (x – 2y) (x – 2y) + 4(x – 2y)

= (x – 2y) (x – 2y + 4)

(23) y2 – xy(1 – x) – x3

= y2 – xy + x2y – x3

= y(y – x) + x2(y – x)

= (y – x) (y + x2)

(24) (ax + by)2 + (bx – ay)2

= [(ax)2 + 2axby + (by)2] + [(bx)2 – 2bxay + (ay)2]

= a2x2 + 2axby + b2y2 + b2x2 – 2axby + a2y2

= a2x2 + b2x2 + a2y2 + b2y2

= x2(a2 + b2) + y2 (a2 + b2)

= (a2 + b2) (x2 + y2)

(25) ab2 + (a – 1) b – 1

= ab2 + ab – b – 1

= ab(b + 1) – 1(b + 1)

= (b + 1) (ab – 1)

(26) x3 – 3x2 + x – 3

= x2(x – 3) + 1 (x – 3)

= (x – 3) (x2 + 1)

(27) ab(x2 + y2) – xy(a2 + b2)

= abx2 + aby2 – a2xy – b2xy

= abx2 – a2xy – b2xy + aby2

= ax (bx – ay) – by (bx – ay)

= (bx – ay) (ax – by)

(28) x2 – x(a + 2b) + 2ab

= x2 – ax – 2bx + 2ab

= x(x – a) – 2b (x – a)

= (x – a) (x – 2b)


3 Comments

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  1. Sir thank you for solving questions

  2. very very thanks uncle.

  3. cool it’s nice

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