Question

(2x – 3y + 4z)2.

Pls answer in expansion form with formula.
The question is from chapter- Expansion
class 9...

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Answer 1

Answer :-

→ 4x² + 9y² + 16z² - 12xy - 24yz + 16zx .

Step-by-step explanation :-

Identity to be used :-

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

We have,

→ ( 2x - 3y + 4z )².

= [ ( 2x ) + ( -3y ) + ( 4z ) ]² .

Here, a = 2x , b = -3y and c = 4z .

$=$ ( 2x )² + ( -3y )² + ( 4z )² + 2(2x)(-3y) + 2(-3y)(4z) + 2(4z)(2x) .

= 4x² + 9y² + 16z² - 12xy - 24yz + 16zx .

Hence, it is solved .

Answer 2

Answer:

4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16zx

Step-by-step explanation:

By using the formula of ( a + b + c )^2.

( a + b + c )^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

(2x – 3y + 4z)^2

=> [(2x) + (-3y) + (4z)]^2

Here suppose, a = 2x, b = -3y, c = 4z

=> (2x)^2 + (-3y)^2 + (4z)^2 + 2 × (2x) × (-3y) + 2 × (-3y) × (4z) + 2 × (4z) × (2x)

=> 4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16zx

.°. (2x – 3y + 4z)^2 = 4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16zx.