Question

Simplify: ${{\left( 3m+5n \right)}^{2}}-{{\left( 2n \right)}^{2}}$.

Answer 1

Hint: We have to use the theorems or identity formulas of ${{\left( a \right)}^{2}}-{{\left( b \right)}^{2}}=\left( a+b \right)\left( a-b \right)$ or ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ to break the given equation. We replace the terms of the theorems to get the values. Using binary operations on the theorems’ result we get the solution of the problem.

Complete step by step answer:
We have been given the subtraction of two square numbers. We use the identity formula of ${{\left( a \right)}^{2}}-{{\left( b \right)}^{2}}=\left( a+b \right)\left( a-b \right)$ to find the simplification of the given equation.
We will replace in the given equation with $a=3m+5n,b=2n$.
$\begin{align}
  & {{\left( 3m+5n \right)}^{2}}-{{\left( 2n \right)}^{2}} \\
 & \Rightarrow \left( 3m+5n+2n \right)\left( 3m+5n-2n \right) \\
\end{align}$
We apply binary operations to find the solution in the form of multiplication.
So, $\left( 3m+5n+2n \right)\left( 3m+5n-2n \right)=\left( 3m+7n \right)\left( 3m+3n \right)=3\left( 3m+7n \right)\left( m+n \right)$.
Now we apply multiplication to find the simplified form
$\begin{align}
  & \Rightarrow 3\left( 3m+7n \right)\left( m+n \right) \\
 & \Rightarrow 3\left( 3{{m}^{2}}+3mn+7mn+7{{n}^{2}} \right) \\
 & \Rightarrow 3\left( 3{{m}^{2}}+10mn+7{{n}^{2}} \right) \\
 & \Rightarrow 9{{m}^{2}}+30mn+21{{n}^{2}} \\
\end{align}$
We can apply the theorem of ${{\left( a+b \right)}^{2}}$, for the part of ${{\left( 3m+5n \right)}^{2}}$.
The identity tells us ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
So, for the replacement we assume $a=3m,b=5n$.
This gives us
$\begin{align}
  & \Rightarrow {{\left( 3m+5n \right)}^{2}} \\
 & \Rightarrow {{\left( 3m \right)}^{2}}+2\times 3m\times 5n+{{\left( 5n \right)}^{2}} \\
 & \Rightarrow 9{{m}^{2}}+30mn+25{{n}^{2}} \\
\end{align}$
Now the given equation is ${{\left( 3m+5n \right)}^{2}}-{{\left( 2n \right)}^{2}}$.
We simplify the rest of the equation to get
$\begin{align}
  & \Rightarrow {{\left( 3m+5n \right)}^{2}}-{{\left( 2n \right)}^{2}} \\
 & \Rightarrow 9{{m}^{2}}+30mn+25{{n}^{2}}-4{{n}^{2}} \\
 & \Rightarrow 9{{m}^{2}}+30mn+21{{n}^{2}} \\
\end{align}$
We get that both the processes give the same outcome.

So, the simplified answer of ${{\left( 3m+5n \right)}^{2}}-{{\left( 2n \right)}^{2}}$ is $9{{m}^{2}}+30mn+21{{n}^{2}}$.

Note: We have used two formulas. The first one is known for the factorisation process. We were able to break the given equation into multiplication of terms. But as we needed a simplified form of the equation, we used the multiplied terms to find the solution. The second theorem is called the value theorem where we try to find the value of the square values only. This process is more direct to find the solution of the problem. Although both theorems are the corollary of one another.