Question

Please answer tomorrow is my exam

Only Q8

Answer 1

Answer:

x^2 + 1/x^2 + 2 - 3x - 3/x

=> x^2 + 1/x^2 + 2 × x × 1/x - 3x - 3/x

=> (x + 1/x) ^2 -3(x + 1/x)

=> (x + 1/x) [ (x+1/x) -3]

Answer 2

Taking LCM as x²

$\frac{ {x}^{4} + 1 +2 {x}^{2} - 3 {x}^{3} - 3x }{ {x}^{2} }$

We can factorise x⁴+1+2x² as (x²+1)² and

-3x³-3x as -3x(x²+1)

So,

$\frac{ {x}^{4} + 1 +2 {x}^{2} - 3 {x}^{3} - 3x }{ {x}^{2} } = \frac{ {( {x}^{2} + 1) }^{2} - 3x( {x}^{2} + 1) }{ {x}^{2} } \\ = \frac{( {x}^{2} + 1) ( {x}^{2} + 1 - 3x)}{ {x}^{2} }$

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