Question

any math expert here?​

Answer 1

Solution:

Given To Factorise –

$\tt = \bigg(x - \dfrac{1}{x} \bigg)^{2} + 3 \bigg(x - \dfrac{1}{x} \bigg) + 2$

Let us assume that,

$\tt \longrightarrow y = x - \dfrac{1}{x}$

So, the polynomial becomes,

$\tt = {y}^{2} + 3y + 2$

By splitting the middle term, we get,

$\tt = {y}^{2} + y + 2y + 2$

$\tt = y(y + 1) + 2(y + 1)$

$\tt = (y+ 2)(y + 1)$

Now, substitute the value of y, we get,

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

Which is our required answer.

Answer 2

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

Step-by-step explanation:

$let \: \: a = (x - \frac{1}{x} )$

now ,

a² + 3a + 2

a² + 2a + a + 2

a(a + 2) + 1(a+2)

(a+1)(a+2)

Now, put value of a

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

stay motivated.