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RD Chapter 5 Factorisation of Algebraic Expressions Ex 5.2 Solutions

Question - 1 : - Factorize each of the following expressions:
p3 + 27

Answer - 1 : -

We know that a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b) (a2 + aft + b2)
p3 + 21 = (p)3 + (3)3
= (p + 3) (p2– p x 3 + 32)
= (p + 3) (p2 – 3p + 9)

Question - 2 : - y3 + 125

Answer - 2 : -

y3 + 125 = (p)3 + (5)3
= (p + 5) (p2 – 5y + 52)
= (P + 5) (p2 – 5y + 25)

Question - 3 : - 1 – 21a3

Answer - 3 : -

1 – 21a3 = (1)3 – (3a)3
= (1 – 3a) [12 + 1 x 3a + (3a)2]
= (1 – 3a) (1 + 3a + 9a2)

Question - 4 : - 8x3y3 + 27a3

Answer - 4 : -

8x3y3 + 27a3
= (2xy + 3a) [(2xy)2 – 2xy x 3a + (3a)2]
= (2xy + 3a) (4x2y – 6xya + 9a2)

Question - 5 : - 64a3 – b3

Answer - 5 : -

64a3 – b3 = (4a)3 –(b)3
= (4a – b) [(4a)2 + 4a x b + (b)2]
= (4a – b) (16a2 + 4ab + b2)

Question - 6 : -

Answer - 6 : -


Question - 7 : - 10x4– 10xy4

Answer - 7 : -

I0x4y- 10xy4 = 10xy(x3 -y3)
= 10xy(x –y) (x2 + xy + y2)

Question - 8 : - 54x6y + 2x3y4

Answer - 8 : -

54 x6y +2x3y4 = 2x3y(27x3 +y3)
= 2x3y[(3x)3 +(y)3]
= 2x3y(3x +y) [(3x)2 -3xx y + y2]
= 2x3y(3x + y) (9x2 -3xy+ y2)

Question - 9 : - 32a3 + 108b3

Answer - 9 : -

32a3 + 108b3
= 4(8a3 + 27b3) = 4 [(2a)3 +(3 b)3]
= 4(2a + 3b) [(2a)2 – 2a x 3b + (3b)2]
= 4(2a + 3b) (4a2 – 6ab + 9b2)

Question - 10 : - (a – 2b)3 – 512b3

Answer - 10 : -

(a – 2b)3 – 512b3
= (a – 2b)3 – (8b)3
= (a – 2b- 8b) [(a – 2b)2 + (a – 2b) x 8b + (8b)2]
= (a – 10b) [a2 + 4b2 – 4ab + 8ab – 16b2 +64b2]
= (a – 10b) (a2 + 4ab + 52b2)

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