Q.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) [14 a - 12 b + 1]2

Solution:

(i) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca 

We can write,

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

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(ii) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca 

We can write,

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

= 4x2 + y2 + z2 - 4xy - 2yz + 4xz

(iii) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca 

We can write,

(-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)Using identity, (x + y + z)= x2 + y2 + z2 + 2xy + 2yz + 2zx 

We can write,

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

We can write,

(-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)Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca 

We can write,

 14a -12b + 12= 14a2 + -12b2+ 12

+2×14a×-12b+2×-12b×1+2×1×14a

=116a2+14b2 + 1 -14ab - b +12a