Factorise the polynomial $$x^4 - 10x^2 + 22$$​ into product of two quadratic polynomials.

Correct option is D

Given:
​$$x^4 - 10x^2 + 22.$$ 
Formula Used:
​$$\text{Roots of equation} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 
Solution:
Let y = x^2
Given Polynomial becomes $$y^2 - 10y + 22.$$​​
$$\text{Roots of equation} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$​
where a = 1, b = -10, and c = 22.
​$$=\frac{ -(-10) \pm \sqrt{ (-10)^2 - 4 \times 1 \times 22}}{2 \times 1} \\ \ \\ = \frac{ (10) \pm \sqrt{ (100) - 88}}{2 } \\ \ \\ = \frac{ (10) \pm \sqrt{ 12}}{2 } \\ \ \\ = \frac{2 (5 \pm \sqrt{ 3})}{2 } \\ \ \\ = 5 \pm \sqrt{ 3}$$​​
Therefore, the two roots of the quadratic are:
y = 5 + √3 and y = 5 - √3
Now, substitute y =$$x^2$$​ back into the factored form:
​$$(x^2 - (5 + √3))(x^2 - (5 - √3))$$​​
The factorisation of $$x^4 - 10x^2 + 22 \ is: (x^2 - (5 + √3))(x^2 - (5 - √3))$$​​