# 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) (7a ^{2} – 63b^{2})**

= 7(a^{2} – 9b^{2})

= 7[(a)^{2} – (3b)^{2}]

= 7 (a – 3b) (a + 3b)

Ans: (d)

**(2) (2x – 32x ^{3})**

= 2x (1 – 16x^{2})

= 2x [(1)^{2} – (4x)^{2}]

= 2x (1 – 4x) (1 + 4x)

Ans: (d)

**(3) x ^{3} – 144x**

= x(x^{2} – 144)

= x[(x)^{2} – (12)^{2}]

= x (x – 12) (x + 12)

Ans: (c)

**(4) (2 – 50x ^{2})**

= 2(1 – 25x^{2})

= 2 [(1)^{2} – (5x)^{2}]

= 2 (1 – 5x) (1 + 5x)

Ans: (d)

**(5) a ^{2} + bc + ab + ac**

= a^{2} + ac + ab + bc

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

= (a + b) (a + c)

Ans: (a)

**(6) pq ^{2} + q(p – 1) – 1**

= pq^{2} + 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) x ^{2} – xz + xy – yz**

= x(x – z) + y(x – z)

= (x + y) (x – z)

Ans: (c)

**(10) (12m ^{2} – 27)**

= 3(4m^{2} – 9)

= 3[(2m)^{2} – (3)^{2}]

= 3(2m – 3) (2m + 3)

Ans: (c)

**(11) x ^{3} – x**

= x (x^{2} – 1)

= x [(x)^{2} – (1)^{2}]

= x (x + 1) (x – 1)

Ans: (d)

**(12) 1 – 2ab – (a ^{2} + b^{2})**

= 1 – 2ab – a^{2} – b^{2}

= 1 – a^{2} – 2ab – b^{2}

= 1 – (a^{2} + 2ab + b^{2})

= (1)^{2} – (a + b)^{2}

= (1 + a + b) (1 – a – b)

Ans: (c)

**(13) x ^{2} + 6x + 8**

= x^{2} + (4 +2)x + 8

= x^{2} + 4x + 2x + 8

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

= (x + 2) (x + 4)

Ans: (c)

**(14) x ^{2} + 4x – 21**

= x^{2} + (7 – 3)x – 21

= x^{2} + 7x – 3x – 21

= x (x + 7) – 3(x + 7)

= (x + 7) (x – 3)

Ans: (b)

**(15) y ^{2} + 2y – 3**

= y^{2} + (3 – 1)y – 3

= y^{2} + 3y – y – 3

= y(y + 3) – 1(y + 3)

= (y – 1) (y + 3)

Ans: (a)

**(16) 40 + 3x – x ^{2}**

= 40 + (8 – 5)x – x^{2}

= 40 + 8x – 5x – x^{2}

= 8 (5 + x) – x(5 + x)

= (5 + x) (8 – x)

Ans: (c)

**(17) 2x ^{2} + 5x + 3**

= 2x^{2} + (2+ 3)x + 3

= 2x^{2} + 2x + 3x + 3

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

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

Ans: (b)

**(18) 6a ^{2} – 13a + 6**

= 6a^{2} – (9 + 4)a + 6

= 6a^{2} – 9a – 4a + 6

= 3a(2a – 3) – 2(2a – 3)

= (3a – 2) (2a – 3)

Ans: (c)

**(19) 4z ^{2} – 8z + 3**

= 4z^{2} – ( 6 + 2)z + 3

= 4z^{2} – 6z – 2z + 3

= 2z(2z – 3) – 1(2z – 3)

= (2z – 1) (2z – 3)

Ans: (a)

**(20) 3 + 23y – 8y ^{2}**

= 3 + (24 – 1)y – 8y^{2}

= 3 + 24y – y – 8y^{2}

= 3(1 + 8y) – y(1 + 8y)

= (1 + 8y) (3 – y)

Ans: (b)