How do you multiply \[{{\left( 2x-3y \right)}^{2}}?\]

Hint: We are asked to multiply \[{{\left( 2x-3y \right)}^{2}}\] and we will start by observing how many terms are there in the given item and then we will try to find a suitable identity to solve this. We will also multiply using the basic approach and we will also solve using the algebraic identity to get the different ways possible to solve the same problem.

Complete step by step answer:
We are given that \[{{\left( 2x-3y \right)}^{2}}\] and as the power is 2, it means that we have to multiply (2x – 3y) with itself. There are various ways to solve the problem. We will start with the basic approach in which we will multiply (2x – 3y) with itself by the multiplication method in which a single term of the first bracket is multiplied by each term of the second bracket. So, we get,
\[\left( 2x-3y \right)\left( 2x-3y \right)=2x\left( 2x-3y \right)-3y\left( 2x-3y \right)\]
Opening the brackets, we get,
\[\Rightarrow \left( 2x-3y \right)\left( 2x-3y \right)=2x\times 2x-2x\times 3y-3y\times 2x-3y\times 3y\]
On simplifying, we get,
\[\Rightarrow \left( 2x-3y \right)\left( 2x-3y \right)=4{{x}^{2}}-6xy-6xy+9{{y}^{2}}\]
Now adding – 6xy – 6xy, we get,
\[\Rightarrow \left( 2x-3y \right)\left( 2x-3y \right)=4{{x}^{2}}-12xy+9{{y}^{2}}\]
Hence, we get,
\[\Rightarrow \left( 2x-3y \right)\left( 2x-3y \right)=4{{x}^{2}}-12xy+9{{y}^{2}}\]
Now, we will work on the other way in which we use the Binomial identities as our problem consists of 2 terms only. We will use \[{{\left( A-B \right)}^{2}}\] and it is given as
\[{{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}\]
So applying \[{{\left( A-B \right)}^{2}}\] on \[{{\left( 2x-3y \right)}^{2}}\] we will consider A as 2x and B as 3y. So, we get,
\[{{\left( 2x-3y \right)}^{2}}={{\left( 2x \right)}^{2}}-2\left( 2x \right)\left( 3y \right)+{{\left( 3y \right)}^{2}}\]
As \[{{\left( 2x \right)}^{2}}=4{{x}^{2}}\] and \[{{\left( 3y \right)}^{2}}=9{{y}^{2}}\]
We get,
\[\Rightarrow {{\left( 2x-3y \right)}^{2}}=4{{x}^{2}}-12xy+9{{y}^{2}}\]
Therefore \[{{\left( 2x-3y \right)}^{2}}=4{{x}^{2}}-12xy+9{{y}^{2}}\] is our solution.

Note: Always remember that when we multiply variables, their power gets added up, that is \[x\times x={{x}^{2}}.\] So, \[x\times x=2x\] cannot be made. Also, remember like terms can only be added, so adding \[4{{x}^{2}}-12y=8{{x}^{2}}y\] is a mistake as \[4{{x}^{2}}\] and 12y are not like terms at all. Also, we need to be careful that \[{{\left( 2x \right)}^{2}}\ne 4x\] as the square is applied to both 2 as well as x. Also, remember when two negatives are added, the term gets added by the sign (–), for example, – 6x – 6x = g = – 12x.