Expand each of the following using suitable identities


Asked by admin @ in Math viewed by 318 People


"Question 4 Expand each of the following, using suitable identities:

(i) (x+2y+4z)^2
(ii) (2x - y + z)^2
(iii) (-2x+3y+2z)^2
(iv) (3a-7b-c)^2
(v) (-2x+5y-3z)^2
(vi) [a/4 - b/2 + 1]^2
Class 9 - Math - Polynomials Page 49"

Answered by admin @



Identity:


An identity is an equality which is true for
all values of a variable in the equality.


(a + b + c)² = a² + b² + c² + 2ab + 2bc
+ 2ca


In an
identity the right hand side expression is called expanded form of the left
hand side expression.

----------------------------------------------------------------------------------------------------


Solution:


i)


(x + 2y +
4z)
²


Using identity,


(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca




Here, a = x, b = 2y and c = 4z




(x + 2y + 4z)
²


 = x² + (2y)² +
(4z)²² + (2×x×2y) + (2×2y×4z) + (2×4z×x)

                   


= x² +
4y
² + 16z²+ 4xy + 16yz + 8xz


 


(ii)  (2x – y + z)²


Using identity,


(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca




Here, a = 2x, b = –y & c
= z




(2x – y + z)
²


 = (2x)² +
(-y)² + z² + (2×2x×-y) + (2×-y×z) +
(2×z×2x)




4x
² + y² + z² –
4xy – 2yz + 4xz



(iii)  (–2x +
3y + 2z)
²


Using identity=


(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca




Here, a = -2x, b = 3y & c
= 2z


(–2x + 3y + 2z)²


(-2x)² + (3y)²+
(2z)
² + (2×-2x×3y) + (2×3y×2z)+ (2×2z×-2x)



4x
² + 9y² + 4z² –
12xy + 12yz – 8xz



(iv)  (3a –
7b – c)
²


Using identity,


(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca




Here, a = 3a, b,= -7b &c = -c



(3a – 7b – c)
²


 (3a)² +
(-7b)
²+ (-c)² + (2×3a×-7b) +
(2×-7b×-c) + (2×-c×3a)



9a
² + 49b² + c² –
42ab + 14bc – 6ac



(v)  (–2x +
5y – 3z)
²


Using identity,


(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca




Here, a = -2x, b = 5y and c = -3z



(–2x + 5y – 3z)
² 


(-2x)² + (5y)² +
(-3z)
² + (2×-2x×5y) + (2×5y×-3z) + (2×-3z×-2x)



4x
² + 25y² + 9z²
20xy – 30yz + 12xz




(vi)  [1/4 a –
1/2 b + 1]
²


Using identity,


(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca




Here, a = a/4 , b = -b/2 
& c = 1


[1/4 a – 1/2 b + 1]²


=(a/4)²+(-b/2 )²+1²+
(2×a/4 ×-b/2)+(2×-b/2 ×1) + (2×1×a/4)



= a²/16+ b²/4+ 1 – ab/4– b + a/2


 


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