Expand ${{\left( 2x+3y \right)}^{2}}$.
Answer
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Hint: The algebraic expression given to us is ${{\left( 2x+3y \right)}^{2}}$ which can be written as: \[\left( 2x+3y \right)\times \left( 2x+3y \right)\]. First of all, we will break the bracket by multiplying $\left( 2x+3y \right)$ with the terms inside the other bracket and we will convert it into four terms having two unknown variables ‘x’ and ‘y’ and after that, we will add the similar terms to get our final result.
Complete step by step solution:
Here, the expression given to us in the question is: ${{\left( 2x+3y \right)}^{2}}$. As we can see, this is an expression in second degree, therefore it will be multiplied to itself twice only. This is because the degree of any mathematical expression decides how many times should the expression be multiplied to itself.
So, at first we will open the bracket by multiplying the value with itself. This can be done as follows:
$\Rightarrow {{\left( 2x+3y \right)}^{2}}=\left( 2x+3y \right)\times \left( 2x+3y \right)$
Now, at first we will multiply ‘2x’ with ‘$\left( 2x+3y \right)$ ’ and then we will multiply ‘3y’ with ‘$\left( 2x+3y \right)$’ and then add both the terms to get our final result. This is done as follows:
$\Rightarrow {{\left( 2x+3y \right)}^{2}}=\left( 2x \right)\times \left( 2x+3y \right)+\left( 3y \right)\times \left( 2x+3y \right)$
Now, on simplifying the above mathematical equation, we get:
$\begin{align}
& \Rightarrow {{\left( 2x+3y \right)}^{2}}=\left( 4{{x}^{2}}+6xy \right)+\left( 6xy+9{{y}^{2}} \right) \\
& \Rightarrow {{\left( 2x+3y \right)}^{2}}=4{{x}^{2}}+6xy+6xy+9{{y}^{2}} \\
& \therefore {{\left( 2x+3y \right)}^{2}}=4{{x}^{2}}+12xy+9{{y}^{2}} \\
\end{align}$
Thus, we get the final result upon simplification as $4{{x}^{2}}+12xy+9{{y}^{2}}$.
Hence, the expansion of ${{\left( 2x+3y \right)}^{2}}$ comes out to be $4{{x}^{2}}+12xy+9{{y}^{2}}$.
Note:
While solving the equation, we should always take care of the signs of different terms. The expression in our problem had only positive terms, so all the terms were positive. But, if the expression had a negative sign then we had to be careful above the negative terms because the answer would differ only in sign.
Complete step by step solution:
Here, the expression given to us in the question is: ${{\left( 2x+3y \right)}^{2}}$. As we can see, this is an expression in second degree, therefore it will be multiplied to itself twice only. This is because the degree of any mathematical expression decides how many times should the expression be multiplied to itself.
So, at first we will open the bracket by multiplying the value with itself. This can be done as follows:
$\Rightarrow {{\left( 2x+3y \right)}^{2}}=\left( 2x+3y \right)\times \left( 2x+3y \right)$
Now, at first we will multiply ‘2x’ with ‘$\left( 2x+3y \right)$ ’ and then we will multiply ‘3y’ with ‘$\left( 2x+3y \right)$’ and then add both the terms to get our final result. This is done as follows:
$\Rightarrow {{\left( 2x+3y \right)}^{2}}=\left( 2x \right)\times \left( 2x+3y \right)+\left( 3y \right)\times \left( 2x+3y \right)$
Now, on simplifying the above mathematical equation, we get:
$\begin{align}
& \Rightarrow {{\left( 2x+3y \right)}^{2}}=\left( 4{{x}^{2}}+6xy \right)+\left( 6xy+9{{y}^{2}} \right) \\
& \Rightarrow {{\left( 2x+3y \right)}^{2}}=4{{x}^{2}}+6xy+6xy+9{{y}^{2}} \\
& \therefore {{\left( 2x+3y \right)}^{2}}=4{{x}^{2}}+12xy+9{{y}^{2}} \\
\end{align}$
Thus, we get the final result upon simplification as $4{{x}^{2}}+12xy+9{{y}^{2}}$.
Hence, the expansion of ${{\left( 2x+3y \right)}^{2}}$ comes out to be $4{{x}^{2}}+12xy+9{{y}^{2}}$.
Note:
While solving the equation, we should always take care of the signs of different terms. The expression in our problem had only positive terms, so all the terms were positive. But, if the expression had a negative sign then we had to be careful above the negative terms because the answer would differ only in sign.
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