Since the coefficients of the equation are all rational numbers, 2+i and 3-\(\sqrt{2}\) are roots, we get 2-i and 3+\(\sqrt{2}\) are also roots of the given equation. Thus (x-(2+i)), (x-(2-i)), (x-(3-\(\sqrt{2}\))) and (x-(3+\(\sqrt{2}\))) are factors. Thus their product.
((x-(2+i))(x-(2-i))(x-(3-\(\sqrt{2}\)))(x-(3+\(\sqrt{2}\))) is a factor of the given polynomial equation.
That is, (x²-4x+5)(x²-6x+7) is a factor. Dividing the given polynomial equation by this factor, we get the other factor as (x²-3x-4) which implies that 4 and −1 are the other two roots. Thus
2+i, 2-i, 3+\(\sqrt{2}\), 3-\(\sqrt{2}\), -1, and 4 are the roots of the given polynomial equation.

Let ∝, β, ૪ be the roots of the equation 
Given \(\frac { \alpha }{ \beta } =\frac { 3 }{ 2 } \Rightarrow 2\alpha =3\beta \Rightarrow \alpha =\frac { 3 }{ 2 } \beta \)
\(\therefore \frac { 3 }{ 2 } \beta ,\beta ,\gamma \) are the roots of the given equation
Then by Vieta's formula,
\(\frac { 3 }{ 2 } \beta +\beta +\gamma =\frac { -b }{ a } =\frac { -(-9) }{ 1 } =9\)
\(\frac { 5 }{ 2 } \beta +\gamma =9\Rightarrow \gamma =9-\frac { 5 }{ 2 } \beta \)
\(\Rightarrow \gamma =\frac { 18-5\beta }{ 2 } ...(2)\)
Also \(\frac { 3 }{ 2 } \beta (\beta )+\beta \gamma +\left( \frac { 3 }{ 2 } \beta \right) \gamma =\frac { c }{ a } =\frac { 14 }{ 1 } =14\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 5 }{ 2 } \beta \left( \frac { 18-5\beta }{ 2 } \right) =14\ [using\ (2)]\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 90\beta }{ 4 } -\frac { 25{ \beta }^{ 2 } }{ 4 } =14\)
Multiplying by \(4,6{ \beta }^{ 2 }+90{ \beta }-25{ \beta }^{ 2 }=56\)
\(19{ \beta }^{ 2 }-90{ \beta }+56=0\)
\(\Rightarrow ({ \beta }-4)(19{ \beta }-14)=0\)
\(\Rightarrow { \beta }=4\)
\({ \beta }=\frac { 14 }{ 19 } \)
When \(\beta\) = 4, the other roots are \(\frac { 3 }{ 2 } (4),4,\frac { 18-5 }{ 2 } (4)\)
\(\Rightarrow 6,4,-1\)

When \(\\ \beta =\frac { 14 }{ 19 } ,\) the other roots are \(\frac { 3 }{ 2 } \beta ,\beta \frac { 18-5\beta }{ 2 } [by(2)]\)
\(\Rightarrow \frac { 3 }{ 2 } \left( \frac { 14 }{ 19 } \right) ,\frac { 14 }{ 19 } ,\frac { 18-5\left( \frac { 14 }{ 19 } \right) }{ 2 } \Rightarrow \frac { 21 }{ 19 } ,\frac { 14 }{ 19 } ,\frac { 136 }{ 19 } \)

Let  α, β and γ  be the roots of x³+ ax²+ bx + c = 0
Then, we get
Σ₁ = α + β + γ = -a ....(1)
Σ₂ = αβ + βγ + γα = b ...(2)
Σ₃ = αβγ = -c ...(3)
We have to form the equation whose roots are  α², β² and γ².
Using (1), (2) and (3), we find the following
Σ₁ = α² + β² + γ² = (α + β + γ )² - 2( αβ + βγ + γα) = (-a)² -2(b) = a²-2b,
Σ₂ = α²β² +  β²γ² + γ²α² = (αβ + βγ + γα)² - 2((αβ)( βγ)(γα) + (γα)(αβ))
= (αβ + βγ + γα)² - 2αβγ (β + γ + α) = (b)² - 2(-c)(-a) = b²-2ca
Σ₃ = α²β²γ² = (αβγ)² = (-c)² = c²
Hence, the required equation is
x³-(α² + β² + γ²)x² + (α²β² + β²γ² + γ²α²)x -  α²β²γ² = 0
That is, x³-(a²-2b)x² + (b²-2ca)x-c² = 0

Given cubic equation 3x²-16x²+23x-6 = 0
Let ∝, \(\frac{1}{\alpha}\) and ૪ be the roots of the equation
[∵ product of two roots is 1]
\((1)\rightarrow { x }^{ 2 }-\frac { 16 }{ 3 } { x }^{ 2 }+\frac { 23 }{ 3 } -2=0 \) ......(1)
comparing  (1) with 
\({ x }^{ 3 }-\left( \frac { \alpha +\beta +\gamma }{ \alpha } \right) +\left( \alpha \frac { 1 }{ \alpha } +\frac { 1 }{ \alpha } .\gamma +\gamma \alpha \right) \)
\(-\alpha \frac { 1 }{ \alpha } .\gamma =0\) ..........(2)
We get, 
\(\alpha +\frac { 1 }{ \alpha } +\gamma =\frac { 16 }{ 3 } \) .......(3)
\(1+\frac { \gamma }{ \alpha } +\gamma \alpha =\frac { 23 }{ 3 } \)
\(\alpha .\frac { 1 }{ \alpha } .\gamma =2\Rightarrow \gamma =2\) ............(4)
Substituting ૪ = 2 in (3)
 \(\alpha +\frac { 1 }{ \alpha } +2=\frac { 16 }{ 3 } \)
\(\Rightarrow \alpha +\frac { 1 }{ \alpha } =\frac { 16 }{ 3 } -2=\frac { 16-6 }{ 3 } =\frac { 10 }{ 3 } \)
\(\Rightarrow \frac { { \alpha }^{ 2 }+1 }{ \alpha } =\frac { 10 }{ 3 } \)

\({ 3x }^{ 2 }+3=10\alpha \)
\({ 3\alpha }^{ 2 }-10\alpha +3=0\)
\(\alpha =\frac { -10 }{ 3 } or\ \alpha =\frac { 1 }{ 3 } \)
\((3\alpha +10)(3\alpha -1)=0\)
\(\alpha =\frac { -10 }{ 3 } \) is not possible \(\Rightarrow \alpha =\frac { 1 }{ 3 } \)
[\(\because \alpha =\frac { -10 }{ 3 } \) will not satisfy(5)]
∴ The roots are 3, \(\frac{1}{3}\), 2.

Given \((\sqrt { 5 } -\sqrt { 3 } )\) is a root
Another root \(\Rightarrow \sqrt { 5 } +\sqrt { 3 } \) 
∴ Sum of the roots \(=\sqrt { 5 } -\sqrt { 3 } +\sqrt { 5 } +\sqrt { 3 } =2\sqrt { 5 } \)
Product of the roots 
\(=(\sqrt { 5 } -\sqrt { 3 } )(\sqrt { 5 } +\sqrt { 3 } )\)
\(=(\sqrt { 5 } )^{ 2 }-{ (\sqrt { 3 } ) }^{ 2 }=5-3=2\)
∴ One of the factor is x² -x (sum of the roots) + product of the roots 
\(\Rightarrow { x }^{ 2 }-2x\sqrt { 5 } +2\)
The other factor also will be \({ x }^{ 2 }-2x\sqrt { 5 } +2\)
\(({ x }^{ 2 }-2x\sqrt { 5 } +2)({ x }^{ 2 }+2x\sqrt { 5 } +2)=0\)
\(\Rightarrow ({ x }^{ 2 }+2-2\sqrt { 5 } x)({ x }^{ 2 }+2+2\sqrt { 5 } x)=0\)
\(\Rightarrow \left( { x }^{ 2 }+2 \right) ^{ 2 }-{ (2\sqrt { 5 } x) }^{ 2 }=0\)
\([\because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 }]\)
\(\Rightarrow { x }^{ 4 }+{ 4x }^{ 2 }+4{ (5)x }^{ 2 }=0\)
\(\Rightarrow { x }^{ 4 }+{ 4x }^{ 2 }+4-{ 20x }^{ 2 }=0\)
\(\Rightarrow { x }^{ 4 }-{ 16x }^{ 2 }+4=0\) is a rational co-efficient polynomial equation.