RS Aggarwal Class 9 Math Chapter 2 Polynomials Exercise 2H Solution

EXERCISE 2H

Question 1:
We know:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (a + 2b + 5c)2
= (a)2 + (2b)2 + (5c)2 + 2(a) (2b) + 2 (2b) (5c) + 2(5c) (a)
= a2 + 4b2 + 25c2 + 4ab + 20bc + 10ac
(ii) (2a – b + c)2
= (2a)2 + (-b)2 + (c)2 + 2 (2a) (-b) + 2(-b) (c) + 2 (c) (2a)
= 4a2 + b2 + c2 – 4ab – 2bc + 4ac.
(iii) (a – 2b – 3c)2
= (a)2 + (-2b)2 + (-3c)2 + 2(a) (-2b) + 2(-2b) (-3c) + 2 (-3c) (a)
= a2 + 4b2 + 9c2 – 4ab + 12bc – 6ac.

Question 2:
We know:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (2a – 5b – 7c)2
= (2a)2 + (-5b)2 + (-7c)2 + 2 (2a) (-5b) + 2 (-5b) (-7c) + 2 (-7c) (2a)
= 4a2 + 25b2 + 49c2 – 20ab + 70bc – 28ac.
(ii) (-3a + 4b – 5c)2
= (-3a)2 + (4b)2 + (-5c)2 + 2 (-3a) (4b) + 2 (4b) (-5c) + 2 (-5c) (-3a)
= 9a2 + 16b2 + 25c2 – 24ab – 40bc + 30ac.
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2h 2.1

Question 3:
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2 (2x) (3y) + 2(3y) (-4z) + 2 (-4z) (2x)
= (2x + 3y – 4z)2

Question 4:
9x2 + 16y2 + 4z2 – 24xy + 16yz – 12xz
= (-3x)2 + (4y)2 + (2z)2 + 2 (-3x) (4y) + 2 (4y) (2z) + 2 (2z) (-3x)
= (-3x + 4y + 2z)2.

Question 5:
25x2 + 4y2 + 9z2 – 20xy – 12yz + 30xz
= (5x)2 + (-2y)2 + (3z)2 + 2(5x) (-2y) + 2(-2y) (3z) + 2(3z) (5x)
= (5x – 2y + 3z)2

Question 6:
(i) (99)2
= (100 – 1)2
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2h 6.1
= (100)2 – 2(100) (1) + (1)2
= 10000 – 200 + 1
= 9801.
(ii) (998)2
= (1000 – 2)2
= (1000)2 – 2 (1000) (2) + (2)2
= 1000000 – 4000 + 4
= 996004.