Question

How do you multiply ${{\left( 2a+3b \right)}^{2}}$?

Answer 1

Hint: The algebraic expression given in the above problem is equal to ${{\left( 2a+3b \right)}^{2}}$. We can see the power of the expression as 2 means we are going to multiply $\left( 2a+3b \right)$ by itself. Then this square expression will look like $\left( 2a+3b \right)\left( 2a+3b \right)$. Now, to multiply these two expressions with a multiply sign we are multiplying $2a$ one by one with the other two terms in the bracket then we multiply $3b$ with other two terms.

Complete step by step solution:
The expression given above is as follows:
${{\left( 2a+3b \right)}^{2}}$
There is a power of 2 so we can expand the above expression by multiplying $\left( 2a+3b \right)$ by itself and we get,
$\Rightarrow \left( 2a+3b \right)\left( 2a+3b \right)$
Now, to multiply $\left( 2a+3b \right)$ by $\left( 2a+3b \right)$ we are going to multiply $2a$ first by $2a$ then add the result of this multiplication to the result of the multiplication of $2a$ by $3b$ then we are going to multiply $3b$ by $2a$ and then add the result of this multiplication $3b$ by $3b$.
$\begin{align}
  & \Rightarrow 2a\left( 2a+3b \right)+3b\left( 2a+3b \right) \\
 & =4{{a}^{2}}+6ab+6ba+9{{b}^{2}} \\
\end{align}$
In the above expression, you can see $ab\And ba$ and we know that in algebra $ab=ba$ so writing $ba=ab$ in the above expression we get,
$=4{{a}^{2}}+6ab+6ab+9{{b}^{2}}$
Now, taking $ab$ as common from $6ab+6ab$ in the above expression we get,
$\begin{align}
  & =4{{a}^{2}}+\left( 6+6 \right)ab+9{{b}^{2}} \\
 & =4{{a}^{2}}+\left( 12 \right)ab+9{{b}^{2}} \\
\end{align}$
Hence, the result of multiplication of the above expression is equal to:
$4{{a}^{2}}+12ab+9{{b}^{2}}$

Note: The alternate way of solving the above expression ${{\left( 2a+3b \right)}^{2}}$ is that as this expression is the square of addition of two terms so we can use the following algebraic identity which is equal to:
${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$
Now, substituting $x=2a$ and $y=3b$ in the above equation we get,
$\begin{align}
  & \Rightarrow {{\left( 2a+3b \right)}^{2}}={{\left( 2a \right)}^{2}}+2\left( 2a \right)\left( 3b \right)+{{\left( 3b \right)}^{2}} \\
 & \Rightarrow {{\left( 2a+3b \right)}^{2}}=4{{a}^{2}}+12ab+9{{b}^{2}} \\
\end{align}$
As you can see that we are getting the same result of multiplication as we that in the above solution.