Question

Hello Mates!

Factorise :

$( {x}^{2} + \frac{1}{{x}^{2}} ) - 4(x + \frac{1}{x} ) + 6$

✔Solution Must
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Have Great future ahead!

Answer 1

$Given : (x +\frac{1}{x})^2 - 4(x +\frac{1}{x}) + 6$

$= > x^2 +\frac{1}{x^2} - 4x -\frac{4}{x} + 6$

$= > x^2 +\frac{1}{x^2} - 2x - 2x -\frac{2}{x} -\frac{2}{x} + 1 + 1 + 4$

$= > x^2 + 1 - 2x + 1 + \frac{1}{x^2} -\frac{2}{x} - 2x -\frac{2}{x} + 4$

$= > x(x +\frac{1}{x} - 2) +\frac{1}{x}(x +\frac{1}{x} - 2) - 2(x +\frac{1}{x} - 2)$

$= > (x +\frac{1}{x} - 2)(x +\frac{1}{x} - 2)$

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

Hope it helps!

Answer 2

=>x2+x21​−4x−x4​+6

= > x^2 +\frac{1}{x^2} - 2x - 2x -\frac{2}{x} -\frac{2}{x} + 1 + 1 + 4=>x2+x21​−2x−2x−x2​−x2​+1+1+4

= > x^2 + 1 - 2x + 1 + \frac{1}{x^2} -\frac{2}{x} - 2x -\frac{2}{x} + 4=>x2+1−2x+1+x21​−x2​−2x−x2​+4

= > x(x +\frac{1}{x} - 2) +\frac{1}{x}(x +\frac{1}{x} - 2) - 2(x +\frac{1}{x} - 2)=>x(x+x1​−2)+x1​(x+x1​−2)−2(x+x1​−2)

= > (x +\frac{1}{x} - 2)(x +\frac{1}{x} - 2)=>(x+x1​−2)(x+x1​−2)

= > (x +\frac{1}{x} - 2)^2=>(x+x1​−2)2