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

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

EXERCISE 7E

OBJECTIVE QUESTIONS

Tick (√) the correct answer in each of the following:

(1) (7a² – 63b²)

= 7(a² – 9b²)

= 7[(a)² – (3b)²]

= 7 (a – 3b) (a + 3b)

Ans: (d)

(2) (2x – 32x³)

= 2x (1 – 16x²)

= 2x [(1)² – (4x)²]

= 2x (1 – 4x) (1 + 4x)

Ans: (d)

(3) x³ – 144x

= x(x² – 144)

= x[(x)² – (12)²]

= x (x – 12) (x + 12)

Ans: (c)

(4) (2 – 50x²)

= 2(1 – 25x²)

= 2 [(1)² – (5x)²]

= 2 (1 – 5x) (1 + 5x)

Ans: (d)

(5) a² + bc + ab + ac

= a² + ac + ab + bc

= a(a + c) + b(a + c)

= (a + b) (a + c)

Ans: (a)

(6) pq² + q(p – 1) – 1

= pq² + qp – q – 1

= pq(q + 1) – 1(q + 1)

= (pq – 1) (q + 1)

Ans: (d)

(7) ab – mn + an – bm

= ab + an – bm – mn

= a(b + n) – m(b + n)

= (a – m) (b + n)

Ans: (b)

(8) ab – a – b + 1

= a(b – 1) – 1(b – 1)

= (a – 1) (b – 1)

Ans: (a)

(9) x² – xz + xy – yz

= x(x – z) + y(x – z)

= (x + y) (x – z)

Ans: (c)

(10) (12m² – 27)

= 3(4m² – 9)

= 3[(2m)² – (3)²]

= 3(2m – 3) (2m + 3)

Ans: (c)

(11) x³ – x

= x (x² – 1)

= x [(x)² – (1)²]

= x (x + 1) (x – 1)

Ans: (d)

(12) 1 – 2ab – (a² + b²)

= 1 – 2ab – a² – b²

= 1 – a² – 2ab – b²

= 1 – (a² + 2ab + b²)

= (1)² – (a + b)²

= (1 + a + b) (1 – a – b)

Ans: (c)

(13) x² + 6x + 8

= x² + (4 +2)x + 8

= x² + 4x + 2x + 8

= x (x + 4) + 2(x + 4)

= (x + 2) (x + 4)

Ans: (c)

(14) x² + 4x – 21

= x² + (7 – 3)x – 21

= x² + 7x – 3x – 21

= x (x + 7) – 3(x + 7)

= (x + 7) (x – 3)

Ans: (b)

(15) y² + 2y – 3

= y² + (3 – 1)y – 3

= y² + 3y – y – 3

= y(y + 3) – 1(y + 3)

= (y – 1) (y + 3)

Ans: (a)

(16) 40 + 3x – x²

= 40 + (8 – 5)x – x²

= 40 + 8x – 5x – x²

= 8 (5 + x) – x(5 + x)

= (5 + x) (8 – x)

Ans: (c)

(17) 2x² + 5x + 3

= 2x² + (2+ 3)x + 3

= 2x² + 2x + 3x + 3

= 2x(x + 1) + 3(x + 1)

= (x + 1) (2x + 3)

Ans: (b)

(18) 6a² – 13a + 6

= 6a² – (9 + 4)a + 6

= 6a² – 9a – 4a + 6

= 3a(2a – 3) – 2(2a – 3)

= (3a – 2) (2a – 3)

Ans: (c)

(19) 4z² – 8z + 3

= 4z² – ( 6 + 2)z + 3

= 4z² – 6z – 2z + 3

= 2z(2z – 3) – 1(2z – 3)

= (2z – 1) (2z – 3)

Ans: (a)

(20) 3 + 23y – 8y²

= 3 + (24 – 1)y – 8y²

= 3 + 24y – y – 8y²

= 3(1 + 8y) – y(1 + 8y)

= (1 + 8y) (3 – y)

Ans: (b)