Question

The expansion of $ {(2x - 3y)^2} $ is:
a) $ 2{x^2} + 3{y^2} + 6xy $
b) $ 4{x^2} + 9{y^2} - 12xy $
c) $ 2{x^2} + 3{y^2} - 6xy $
d) $ 4{x^2} + 9{y^2} + 12xy $

Answer 1

Hint: The given algebraic expression is $ {(2x - 3y)^2} = (2x - 3y) \times (2x - 3y) $ . First of all we will break the bracket by multiplying $ (2x - 3y) $ with the terms inside the other bracket and we will convert it into four terms having two unknown variables that is $ x,y $ and after that we will add the similar terms and we will have a final value.

Complete step-by-step answer:
Here, the expression given in the question is:
 $ {(2x - 3y)^2} $
So, at first we will open the bracket by multiplying the value with itself:
 $ \Rightarrow {(2x - 3y)^2} = (2x - 3y) \times (2x - 3y) $
So now at first we will multiply $ 2x $ with $ (2x - 3y) $ then we will multiply $ 3x $ with $ (2x - 3y) $ and we will subtract both of them.
 So, the resulting values are:
 $ \Rightarrow 2x \times (2x - 3y) = 4{x^2} - 6xy $ -------(1)
And,
 $ \Rightarrow 3y(2x - 3y) = 6xy - 9{y^2} $ ------(2)
 Now we need to subtract (2) from (1) and we will obtain
 \[\Rightarrow 4{x^2} - 6xy - 6xy + 9{y^2} = 4{x^2} - 12xy + 9{y^2}\] ------(3)
The above equation is the simplified version of the expression $ {(2x - 3y)^2} $
So the final answer we got is \[6{x^2} - 12xy + 9{y^2}\]
Therefore, the simplified version of the expression $ {(2x - 3y)^2} $ is \[4{x^2} - 12xy + 9{y^2}\]
So, the correct answer is “Option B”.

Note: While solving the equation we should take care of the sign or else we may get another value. And we take care while putting the value of the variable. Keep all the variables of the expression in a proper arrangement to avoid errors.
An alternative way of doing this is to use the binomial formula which says $ {(a - b)^2} = {a^2} - 2ab + {b^2} $ , in our case $ a = 2x $ and $ b = 3y $ so using this formula we have:
 $ \Rightarrow {(2x - 3y)^2} = {(2x)^2} - 2(2x)(3y) + {(3y)^2} $
Which on simplification gives us:
 $ \Rightarrow {(2x - 3y)^2} = 4{x^2} - 12xy + 9{y^2} $ and this result is the same as we obtained earlier