Question

Factorise the algebraic expression ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3$
(a) $\left( x+\dfrac{1}{x}+1 \right)\left( 2x-\dfrac{1}{x}-4 \right)$
(b) $\left( x-\dfrac{1}{x}-1 \right)\left( x+\dfrac{1}{x}-2 \right)$
(c) $\left( x-\dfrac{1}{x}-1 \right)\left( x-\dfrac{1}{x}+1 \right)$
(d) $\left( x+\dfrac{1}{x}-1 \right)\left( 2x-\dfrac{1}{x}-5 \right)$

Answer 1

Hint: Factorization of an expression can be done through different methods. One of the methods is to factorize the expression by making it a perfect square.
For example:
$\begin{align}
  & {{x}^{2}}+2x \\
 & ={{x}^{2}}+2x+1-1 \\
 & ={{\left( x+1 \right)}^{2}}-{{\left( 1 \right)}^{2}} \\
 & =\left( x+1+1 \right)\left( x+1-1 \right) \\
 & =x\left( x+2 \right) \\
\end{align}$
The following formulas would be helpful while solving these types of questions:
$\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \\
\end{align}$

Complete step by step solution:
We have the given expression as ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3$. Let us see how we can simplify it and make its factors.
Let us assume the following:
$\begin{align}
  & {{x}^{2}}=y \\
 & \dfrac{1}{{{x}^{2}}}=z \\
\end{align}$
Now, on carefully observing the two variables y and z, we find that their product is a constant value 1, and we also have a constant value of -3 in the expression already present. So, we can try to make a perfect square of the kind of ${{\left( x+\dfrac{1}{x} \right)}^{2}}+k$ , where k is some constant that we can find out.
$\begin{align}
  & {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-3 \\
 & ={{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2-1 \\
 & ={{\left( x \right)}^{2}}+\left( \dfrac{1}{x} \right)^2 -2\times x\times \dfrac{1}{x}-1 \\
 & ={{\left( x+\dfrac{1}{x} \right)}^{2}}-1 \\
\end{align}$
Here, we have split -3 into the sum of -1 and -2 which made it easier to form a perfect square in terms of x. now, we can further simplify this using the identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$,
$\begin{align}
  & {{\left( x+\dfrac{1}{x} \right)}^{2}}-{{\left( 1 \right)}^{2}} \\
 & =\left( x+\dfrac{1}{x}-1 \right)\left( x+\dfrac{1}{x}+1 \right) \\
\end{align}$
Hence, the factored expression is given as $\left( x+\dfrac{1}{x}-1 \right)\left( x+\dfrac{1}{x}+1 \right)$. From the options, option (c) matches with the required answer. So, option (c) is the correct answer.

Note: Here, one must take care of the signs while simplifying the given expression. Like we split -3 into sum of -1 and -2, there might be chances that one might ignore the ‘- ‘sign present in front of 3 and may write it as 1+2. This would lead to estimation of wrong answers and the new expression might also become non-factorizable. One must also have a great hold on the identities and their expansions as their application only allows us to factorize these types of expressions.