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 – ax – bx + ab
= x (x – a) – b (x – a)
= (x – a) (x – b)
(19) ab2 – bc2 – ab + 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)
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