Question

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) [ $\frac{1 }{ 4}$ a - $\frac{1 }{ 2}$ b + 1]2

Answer 1

It is known that,

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

(i) (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) (2x – y + z) 2 = (2x)2 + (-y)2 + (z)2 + 2 (2x)(-y) + 2 (-y) (z) + 2 (z)(2x)

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

(iii) (–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) (–2x + 3y + 2z) 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 = (-2x)2 + (5y)2 + (-3z)2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)

= 4x2 + 25y2 + 9z2 - 20xy - 30yz + 12xz

(vi) [$\frac{1 }{ 4}$ a - $\frac{1 }{ 2}$2 b + 1]2 = ($\frac{1 }{ 4}$ a)2 + (-$\frac{1 }{ 2}$b)2 + (1)2 + 2($\frac{1 }{ 4}$ a)(-$\frac{1 }{ 2}$ b) + 2 (-$\frac{1 }{ 2}$ b)(1) + 2 ($\frac{1 }{ 4}$ a) (1)

= $\frac{1 }{ 16}$ a2 + $\frac{1 }{ 4}$ b2 + 1 - $\frac{1 }{ 4}$ ab - b + $\frac{1 }{ 2}$ a