Question

"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"

Answer 1

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.

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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 + 4y² + 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)

= 4x² + 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)

= 4x² + 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)

= 9a² + 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)

= 4x² + 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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Answer 2

Answer:

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Step-by-step explanation:

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.

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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 + 4y² + 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)

= 4x² + 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)

= 4x² + 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)

= 9a² + 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)

= 4x² + 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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