NCERT Solutions
Class 9 Maths
Polynomials

Ex.2.5 Q.4
Expand each of the following, using suitable identities:
(1) (x + 2y + 4z)2
(2) (2x – y + z)2
(3) (–2x + 3y + 2z)2
(4) (3a – 7b – c)2
(5) (–2x + 5y – 3z)2
(6) [ –
+ 1]2
(1) (x + 2y + 4z)2
= x2 + (2y)2 + (4z)2 + 2 × x × 2y + 2 × 2y × 4z + 2 × 4z × x
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx
(2) (2x – y + z)2
= (2x)2 + (-y)2 + z2 + 2 × 2x × (-y) + 2 × (-y) × z + 2 × 2x × z
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 4x2 + y2 + z2 – 4xy – 2yz + 4xz
(3) (–2x + 3y + 2z)2
= (-2x)2 + (3y)2 + (2z)2 + 2 × (-2x) × 3y + 2 × 3y × 2z + 2 × (-2x) × 2z
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 4x2 + 9y2 + 4z2 – 12xy + 12yz - 8xz
(4) (3a – 7b – c)2
= (3a)2 + (-7b)2 + (-c)2 + 2 × 3a × (-7b) + 2 × (-7b) × (-c) + 2 × 3a × (-c)
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca
(5) (–2x + 5y – 3z)2
= (-2x)2 + (5y)2 + (-3z)2 + 2 × (-2x) × 5y + 2 × 5y × (-3z) + 2 × (-2x) × (-3z)
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 4x2 + 25y2 + 9z2 – 20xy - 30yz + 12xz
(6) [ –
+ 1]2
= ( )2 + (-
)2 + 12 + 2 × (
) × (-
) + 2 × (-
) × 1 + 2 × 1 × (
)
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= +
+ 1 –
– b +