How do you factor \[144{{x}^{2}}-81\]?

Hint: This question is from the topic of algebra. For solving this question, we should know how to find the prime factorization of any number. In this question, we will first find the prime factorization of 144. After that, we will find the prime factorization of 81. After that, we will use the formula of \[{{a}^{2}}-{{b}^{2}}\] which is equal to \[\left( a+b \right)\left( a-b \right)\]. After solving further processes, we will get our answer. After that, we will see alternate methods to solve this question.

Complete step by step answer:
Let us solve this question.
This question has asked us to find the factor of the term which is given in the question.
The term given in the question is
\[144{{x}^{2}}-81\]
Let us first find out the prime factorization of 144.
\[\begin{align}
  & 2\left| \!{\underline {\,
  144 \,}} \right. \\
 & 2\left| \!{\underline {\,
  72 \,}} \right. \\
 & 2\left| \!{\underline {\,
  36 \,}} \right. \\
 & 2\left| \!{\underline {\,
  18 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
Hence, prime factorization of 144 will be \[2\times 2\times 2\times 2\times 3\times 3\times 1\]
Now, let us find the prime factorization of 81.
\[\begin{align}
  & 3\left| \!{\underline {\,
  81 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
Hence, prime factorization of 81 will be \[3\times 3\times 3\times 3\times 1\].
So, we can write
\[144{{x}^{2}}-81=2\times 2\times 2\times 2\times 3\times 3\times 1\times {{x}^{2}}-3\times 3\times 3\times 3\times 1\]
Now, taking common \[3\times 3\times 1\] in the above equation, we can write the above equation as
\[\Rightarrow 144{{x}^{2}}-81=3\times 3\times 1\times \left( 2\times 2\times 2\times 2\times {{x}^{2}}-3\times 3 \right)\]
Now, we can write the above equation as
\[\Rightarrow 144{{x}^{2}}-81=9\left( 2\times 2\times 2\times 2\times {{x}^{2}}-3\times 3 \right)\]
\[\Rightarrow 144{{x}^{2}}-81=9\left( 4\times 4\times {{x}^{2}}-3\times 3 \right)\]
The above equation can also be written as
\[\Rightarrow 144{{x}^{2}}-81=9\left( {{4}^{2}}\times {{x}^{2}}-{{3}^{2}} \right)\]
The above equation can also be written as
\[\Rightarrow 144{{x}^{2}}-81=9\left( {{\left( 4x \right)}^{2}}-{{3}^{2}} \right)\]
Using the formula \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we can write the above equation as
\[\Rightarrow 144{{x}^{2}}-81=9\left( \left( 4x \right)+3 \right)\left( \left( 4x \right)-3 \right)\]
\[\Rightarrow 144{{x}^{2}}-81=9\left( 4x+3 \right)\left( 4x-3 \right)\]

Hence, the factor of the term \[144{{x}^{2}}-81\] is \[9\left( 4x+3 \right)\left( 4x-3 \right)\]

Note: For solving this type of question easily, we should know how to find prime factorization of any number. Remember the following formula:
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
We can solve this question by alternate method.
The term which we have to solve is
\[144{{x}^{2}}-81\]
As we know that 144 can be written as 16 multiplied by 9 and 81 can be written as 9 multiplied by 9, so we can the above term as
\[\Rightarrow 144{{x}^{2}}-81=16\times 9{{x}^{2}}-9\times 9\]
Now, taking 9 as common in the above equation, we get
\[\Rightarrow 144{{x}^{2}}-81=9\left( 16{{x}^{2}}-9 \right)\]
Now, using the formula \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we can write
\[\Rightarrow 144{{x}^{2}}-81=9\left( {{\left( 4x \right)}^{2}}-{{3}^{2}} \right)=9\left( 4x+3 \right)\left( 4x-3 \right)\]
As we have got the same from this method, so we can use this method too.