Question

What is the product of $\left( 2x+3 \right)$ and $\left( 2x-3 \right)$ ?

Answer 1

Hint: To obtain the product of two terms we will use a special product of binomial methods. Firstly we will write the formula to solve the special product of binomials then we will compare it by our two terms and get the value. Then we will simplify it further and get the desired answer.

Complete step by step solution:
We have to find the product of $\left( 2x+3 \right)$ and $\left( 2x-3 \right)$so we can write,
\[\left( 2x+3 \right)\left( 2x-3 \right)\]……$\left( 1 \right)$
We have the formula for special product of binomial as below:
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$…….$\left( 2 \right)$
On comparing equation (1) with equation (2) left hand side we get,
$\begin{align}
  & a=2x \\
 & b=3 \\
\end{align}$
On substituting the above value in equation (2) and simplifying it we get,
\[\begin{align}
  & \Rightarrow \left( 2x+3 \right)\left( 2x-3 \right)={{\left( 2x \right)}^{2}}-{{\left( 3 \right)}^{2}} \\
 & \therefore \left( 2x+3 \right)\left( 2x-3 \right)=4{{x}^{2}}-9 \\
\end{align}\]
So we get the answer as $4{{x}^{2}}-9$

Hence the product of $\left( 2x+3 \right)$ and $\left( 2x-3 \right)$ is $4{{x}^{2}}-9$.

Note: The two binomials $\left( a+b \right)$ and $\left( a-b \right)$ are special as they are formed by the same terms only the sign in each binomial is different. The product of such binomials is equal to the difference of squares of the terms i.e. $a,b$. We can use simple multiplication of each term in bracket one to each term in bracket two for finding the answer and can see that we will get the same answer. Another method if the formula is not known is as follows:
We have to find product of
$\left( 2x+3 \right)$ and $\left( 2x-3 \right)$
So we will multiply them as follows:
\[\begin{align}
  & \left( 2x+3 \right)\left( 2x-3 \right) \\
 & \Rightarrow \left( 2x\times 2x \right)+\left( 2x\times -3 \right)+\left( 3\times 2x \right)+\left( 3\times -3 \right) \\
 & \Rightarrow 4{{x}^{2}}-6x+6x-9 \\
 & \Rightarrow 4{{x}^{2}}-9 \\
\end{align}\]
We got the same answer as above.
The formula method is most preferred because in case the terms are big then we can reduce the calculation part and solve it easily.