Question

factorise x^2 + 1/x^2-7(x-1/x)+8

Answer 1

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Answer 2

The factorization of $x^2+\dfrac{1}{x^2} -7(x-\dfrac{1}{x})+8$ is

$[(x-\dfrac{1}{x})-2][(x-\dfrac{1}{x})-5]$.

Step-by-step explanation:

We have,

$x^2+\dfrac{1}{x^2} -7(x-\dfrac{1}{x})+8$      .......(1)

To find, the factorization of $x^2+\dfrac{1}{x^2} -7(x-\dfrac{1}{x})+8=?$

We know that,

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

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

⇒ $x^2+\dfrac{1}{x^2}=(x-\dfrac{1}{x})^{2}+2$     ......(2)

From (1) and (2), we get

$(x-\dfrac{1}{x})^{2}+2-7(x-\dfrac{1}{x})+8$

$=(x-\dfrac{1}{x})^{2}-7(x-\dfrac{1}{x})+10$

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

$a^{2}-7a+10$

$=a^{2}-7a-2a+10$

$=(a-2)(a-5)$

Put $a=x-\dfrac{1}{x}$

$[(x-\dfrac{1}{x})-2][(x-\dfrac{1}{x})-5]$

Hence, the factorization of $x^2+\dfrac{1}{x^2} -7(x-\dfrac{1}{x})+8$ is

$[(x-\dfrac{1}{x})-2][(x-\dfrac{1}{x})-5]$.