Question

$(x ^{2} + \frac{1}{x ^{2} } ) - 4(x - \frac{1}{x} ) + 6$
factorize this expression​

Answer 1

Step-by-step explanation:

Corrected Question :-

Factorise [x²+(1/x²)] -4[x+(1/x)] +6

Solution :-

Given expression is

[x²+(1/x²)] -4[x+(1/x)] +6 ----(1)

We know that

(a+b)² = a²+2ab+b²

=> a²+b² = (a+b)²-2ab

Therefore, x²+(1/x)² = [x+(1/x)]² - 2(x)(1/x)

=> x²+(1/x)² = [x+(1/x)]² -2 (x/x)

=> x²+(1/x)² = [x+(1/x)]² - 2(1)

=> x²+(1/x)² = [x+(1/x)]² - 2 ----------(2)

Now, equation (1) becomes

[x+(1/x)]²-2-4[x+(1/x)] +6

=> [x+(1/x)]²-4[x+(1/x)]+(6-2)

=> [x+(1/x)]²-4[x+(1/x)]+4

Put x+(1/x) = a then

We have, a²-4a+4

=> (a)²-2(a)(2)+(2)²

=> (a-2)²

Since , (a-b)² = a²-2ab+b²

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

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

Answer :-

[x²+(1/x²)] -4[x+(1/x)] +6 =

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

Used formulae:-

→(a+b)² = a²+2ab+b²

→ (a-b)² = (a+b)²-4ab

→ (a+b)² = (a-b)² +4ab

→ (a-b)² = a²-2ab+b²

Note :-

If we solve the given expression in the question then we get imaginary roots in the factorization.

Answer 2

Correct Question :

• $\sf{[x^2\:+\:({\dfrac{1}{x^2}})]\:-\:4\:[x\:+\:({\dfrac{1}{x}})]\:+\:6}$

Solution :

• $\bold{[x\:+\:{\dfrac{1}{x}}\:-\:2]^2}$

Step-by-step explanation :

$\implies\sf{[x^2\:+\:({\dfrac{1}{x^2}})]\:-\:4\:[x\:+\:({\dfrac{1}{x}})]\:+\:6}$

Taking common square,

$\implies\sf{[x\:+\:({\dfrac{1}{x}})]^2\:-\:4\:[x\:+\:({\dfrac{1}{x}})]\:+\:6}$

Using Identity,

${\boxed{\bullet\:{(a\:+\:b)^2\:=\:a^2\:+\:2ab\:+\:b^2}}}$

where,

• $\sf{a\:=\:x}$
• $\sf{b\:=\:{\dfrac{1}{x}}}$

Substituting values in Formula, we get,

$\implies\sf{[x^2\:+\:2\:(x)\:({\dfrac{1}{x}})\:+\:{\dfrac{1}{x}}^2]\:-\:4\:[x\:+\:({\dfrac{1}{x}})]\:+\:6}$

$\implies\sf{(x\:-\:{\dfrac{1}{x}})^2\:-\:2\:\times\:1\:-\:4\:[x\:+\:({\dfrac{1}{x}})]\:+\:6}$

$\implies\sf{(x\:-\:{\dfrac{1}{x}})^2\:-\:2\:-\:4\:[x\:+\:({\dfrac{1}{x}})]\:+\:6}$

Taking (-2) from LHS to RHS, we get,

$\implies\sf{(x\:-\:{\dfrac{1}{x}})^2\:-\:4\:[x\:+\:({\dfrac{1}{x}})]\:+\:6\:-\:2}$

$\implies\sf{(x\:-\:{\dfrac{1}{x}})^2\:-\:4\:[x\:+\:({\dfrac{1}{x}})]\:+\:4}$

Using identity,

${\boxed{\bullet\:{a^2\:+\:b^2\:-2ab\:=\:(a\:-\:b)^2}}}$

where,

• $\sf{a\:=\:x\:-\:{\dfrac{1}{x}}}$
• $\sf{b\:=\:2}$

Substituting values in Formula, we get,

$\implies\sf{[x\:+\:{\dfrac{1}{x}}\:-\:2]^2}$

Hence, the answer is $\bold{[x\:+\:{\dfrac{1}{x}}\:-\:2]^2}$.