Question

Factorise using any of the methods (splitting the middle term, grouping, using identities)
$x^2 + \frac{1}{x^2} - 2-3x +\frac{3}{x}$

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:x^2 + \dfrac{1}{x^2} - 2-3x +\dfrac{3}{x}$

can be rewritten as by Regrouping, we get

$\rm \: = \: \: \bigg(x^2 + \dfrac{1}{x^2} - 2 \bigg)- \bigg(3x - \dfrac{3}{x} \bigg)$

$\rm \: = \: \: \bigg(x^2 + \dfrac{1}{x^2} - 2 \bigg)-3 \bigg(x - \dfrac{1}{x} \bigg)$

$\rm \: = \: \: \bigg(x^2 + \dfrac{1}{x^2} - 2 \times 1 \bigg)-3 \bigg(x - \dfrac{1}{x} \bigg)$

$\rm \: = \: \: \bigg(x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x} \bigg)-3 \bigg(x - \dfrac{1}{x} \bigg)$

We know,

$\boxed{ \rm \: {x}^{2} + {y}^{2} - 2xy = {(x - y)}^{2} }$

So, using this identity,

$\rm \: = \: \: {\bigg(x - \dfrac{1}{x} \bigg) }^{2} - 3\bigg(x - \dfrac{1}{x} \bigg)$

Now, take out common from these, we get

$\rm \: = \: \: {\bigg(x - \dfrac{1}{x} \bigg) } \bigg \{\bigg(x - \dfrac{1}{x} \bigg) - 3 \bigg \}$

$\rm \: = \: \: {\bigg(x - \dfrac{1}{x} \bigg) } \bigg \{x - \dfrac{1}{x} - 3 \bigg \}$

Basic Concept Used :-

Factorisation by Regrouping Terms

Sometimes it happens that it is not easy to take out common term in the expressions then

We have to make the groups of the terms.

Then choose the common factor among these groups.

Find the common factor and it will give the required factors.

Additional Information :-

1. Method of Common Factors

In this method, we have to write the irreducible factors of all the terms

Then find the common factors amongst all these irreducible factors.

The required is the product of the common term we had taken out and the remaining terms.

2. More Identities to know :-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)