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RD Chapter 5- Factorisation of Algebraic Expressions Ex-5.2 Interview Questions Answers

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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