Question

Factorize it!
Need answer with explanation
$( {x}^{2} + \frac{1}{ {x}^{2} } ) - 4(x + \frac{1}{x} ) + 6$
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Answer 1

Answer:

$\bigg( x + \dfrac{1}{x} - 2\bigg)^2$

Step-by-step explanation:

Given equation is $\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) - 4\bigg(x + \dfrac{1}{x} \bigg) + 6$.

$\implies x^2 + \dfrac{1}{x^2} + 6 - 4\bigg( x + \dfrac{1}{x}\bigg) \\\\\\\implies x^2 + \dfrac{1}{x^2} + ( 2 + 4 ) - 4\bigg( x + \dfrac{1}{x}\bigg) \\\\\\\implies x^2 + \dfrac{1}{x^2} + 2 + 4 - \bigg( x + \dfrac{1}{x}\bigg) \\\\\\\implies x^2 + \dfrac{1}{x^2} + 2\bigg( x \times \dfrac{1}{x} \bigg) + 4 - 4\bigg( x + \dfrac{1}{x}\bigg)$

$\implies \bigg( x + \dfrac{1}{x}\bigg)^2 + 4 - 4\bigg( x + \dfrac{1}{x}\bigg)$

Let $x + \dfrac{1}{x}=a$

= > a^2 + 4 - 4a

= > a^2 - 4a + 4

= > a^2 - ( 2 + 2 )a + 4

= > a^2 - 2a - 2a + 4

= > a( a - 2 ) - 2( a - 2 )

= > ( a - 2 )( a - 2 )

= > ( a - 2 )^2

= > $\bigg( x + \dfrac{1}{x} - 2\bigg)^2$

Hence the required factorised form of the given equation is $\bigg( x + \dfrac{1}{x} - 2\bigg)^2$

Answer 2

Answer:

(x + 1/x - 2)(x + 1/x - 2)

Step-by-step explanation:

Given,

Equation = (x² + 1/x²) - 4 (x + 1/x) + 6

=> (x² + 1/x²) - 4 (x + 1/x) + 6

=> x² + 1/x² + 6 - 4 (x + 1/x)

=> x² + 1/x² + (2 + 4) - 4 (x + 1/x)

=> x² + 1/x² + 2 + 4 - 4 (x + 1/x)

=> x² + 1/x² + 2(x × 1/x) + 4 - 4 (x + 1/x)

=> (x + 1/x)² + 4 - 4 (x + 1/x)

Let x + 1/x = a

=> a² + 4 - 4a

=> a² - 4a + 4

=> a² - (2 + 2)a + 4

=> a² - 2a - 2a + 4

=> a(a - 2) - 2(a - 2)

=> (a - 2)²

Now, putting the given values of 'a'.

=> (a - 2)²

=> (x + 1/x - 2)²

=> (x + 1/x - 2)(x + 1/x - 2)