## RS Aggarwal Class 8 Math Seventh Chapter Factorisation Exercise 7B Solution

## EXERCISE 7B

#### Factorise:

### FORMULA: (a^{2} – b^{2}) = (a + b) (a – b)

**(1) x ^{2} – 36**

= x^{2} – 6^{2}

= (x – 6) (x + 6)

**(2) 4a ^{2} – 9**

= (2a)^{2} – (3)^{2}

= (2a – 3) (2a + 3)

**(3) 81 – 49x ^{2}**

= (9)^{2} – (7x)^{2}

= (9 – 7x) (9 + 7x)

**(4) 4x ^{2} – 9y^{2}**

= (2x)^{2} – (3y)^{2}

= (2x – 3y) (2x + 3y)

**(5) 16a ^{2} – 225b^{2}**

= (4a)^{2} – (15b)^{2}

= (4a – 15b) (4a + 15b)

**(6) 9a ^{2}b^{2} – 25**

= (3ab)^{2} – (5)^{2}

= (3ab – 5) (3ab + 5)

**(7) 16a ^{2} – 144**

= (4a)^{2} – (12)^{2}

= (4a – 12) (4a + 12)

= [4(a – 3)] [4(a + 3)]

= (4 × 4) (a – 3) (a + 3)

= 16 (a – 3) (a + 3)

**(8) 63a ^{2} – 112b^{2}**

= 7(9a^{2} – 16b^{2})

= 7 [(3a)^{2} – (4b)^{2}]

= 7 (3a – 4b) (3a + 4b)

**(9) 20a ^{2} – 45b^{2}**

= 5(4a^{2} – 9b^{2})

= 5 [(2a)^{2} – (3b)^{2}]

= 5 (2a – 3b) (2a + 3b)

**(10) 12x ^{2} – 27**

= 3(4x^{2} – 9)

= 3[(2x)^{2} – (3)^{2}]

= 3 (2x – 3) (2x + 3)

**(11) x ^{3} – 64x**

= x[(x)^{2} – (8)^{2}]

= x (x – 8) (x + 8)

**(12) 16x ^{5} – 144x^{3}**

= 16x^{3} (x^{2} – 9)

= 16x^{3} [(x)^{2} – (3)^{2}]

= 16x^{3} (x – 3) (x + 3)

**(13) 3x ^{5} – 48x^{3}**

= 3x^{3}[(x)^{2} – 16]

= 3x^{3}[(x)^{2} – (4)^{2}]

= 3x^{3} (x – 4) (x + 4)

**(14) 16p3 – 4p**

= 4p (4p^{2} – 1)

= 4p [(2p)^{2} – (1)^{2}]

= 4p (2p – 1) (2p + 1)

**(15) 63a ^{2}b^{2} – 7**

= 7 (9a^{2}b^{2} – 1)

= 7 [(3ab)^{2} – (1)^{2}]

= 7 (3ab – 1) (3ab + 1)

**(16) 1 – (b – c) ^{2}**

= (1)^{2} – (b – c)^{2}

= (1 – b + c) (1 + b – c)

**(17) (2a + 3b) ^{2} – 16c^{2}**

= (2a + 3b)^{2} – (4c)^{2}

= (2a + 3b – 4c) (2a + 3b + 4c)

**(18) (l + m) ^{2} – (l – m)^{2}**

= (l + m + l – m) (l + m – l + m)

= 2l × 2m = 4lm

**(19) (2x + 5y) ^{2} – 1**

= (2x + 5y)^{2} – (1)^{2}

= (2x + 5y – 1) (2x + 5y + 1)

**(20) 36c ^{2} – (5a + b)^{2}**

= (6c)^{2} – (5a + b)^{2}

= (6c + 5a + b) (6c – 5a – b)

**(21) (3x – 4y) ^{2} – 25z^{2}**

= (3x – 4y)^{2} – (5z)^{2}

= (3x – 4y + 5z) (3x – 4y – 5z)

**(22) x ^{2} – y^{2} – 2y – 1**

= x^{2} – (y^{2} + 2y + 1)

= x^{2} – (y + 1)^{2}

= (x + y + 1) (x – y – 1)

**(23) 25 – a ^{2} – b^{2} – 2ab**

= 25 – (a^{2} + 2ab + b^{2})

= (5)^{2 }– (a + b)^{2}

= (5 + a + b) (5 – a – b)

**(24) 25a ^{2} – 4b^{2} + 28bc – 49c^{2}**

= 25a^{2} – [(2b)^{2} – (2 × 2b × 7c) + (7c)^{2}]

= (5a)^{2} – (2b – 7c)^{2}

= (5a + 2b – 7c) (5a – 2b + 7c)

**(25) 9a ^{2} – b^{2} + 4b – 4**

= 9a^{2} – [(b)^{2} – (2 × b × 2) + (2)^{2}]

= (3a)^{2} – (b – 2)^{2}

= (3a + b – 2) (3a – b + 2)

**(26) 100 – (x – 5) ^{2}**

= (10)^{2} – (x – 5)^{2}

= (10 + x – 5) (10 – x + 5)

= (5 + x) (15 – x)

**(27) Evaluate {(405) ^{2 }– (395)^{2}}**

= (405 + 395) (405 – 395)

= 800 × 10 = 8000

**(28) Evaluate {(7.8) ^{2} – (2.2)^{2}}**

= (7.8 + 2.2) (7.8 – 2.2)

= 10 × 5.6 = 56