The roots of x³+2x²+3x+4 = 0 are ∝, β, ૪
∴ ∝ + β + ૪ = -co-efficient of x² = -2   ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3   ...(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4      ...(3)
Form the equation whose roots are ∝-β-૪
∴ -∝-β-૪ = -(∝+β+૪)
= -(-2) = 2
∝β + β૪ + ૪∝ = 3
(-∝)(-β)(-૪) = -(∝β૪) = -(-4) = 4
∴ The required cubic equation is 
x³-(-∝-β-૪)x²+(∝β+β૪+૪∝)
x-[(-∝)(-β)(-૪)] = 0
⇒ x³-(2)x²+3x-4 = 0
⇒ x³-2x²+3x-4 = 0

The roots of x³+2x²+3x+4 = 0 are ∝, β, ૪
∴ ∝ + β + ૪ = -co-efficient of x² = -2   .........(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3   ..........(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4  .......(3)
From the cubic equation whose roots are \(\frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } ,\frac { 1 }{ \gamma } \)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } =\frac { \beta \gamma +\gamma \alpha +\alpha \beta }{ \alpha \beta \gamma } =\frac { 3 }{ -4 } =\frac { -3 }{ 4 } \)
\(\frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } =\frac { \gamma +\alpha +\beta }{ \alpha \beta \gamma } =\frac { -2 }{ -4 } =\frac { 1 }{ 2 } \)
\(\left( \frac { 1 }{ \alpha } \right) \left( \frac { 1 }{ \beta } \right) \left( \frac { 1 }{ \gamma } \right) =\frac { 1 }{ \alpha \beta \gamma } =\frac { 1 }{ -4 } =-\frac { 1 }{ 4 } \)
∴ The required cubic equation is 
\({ x }^{ 3 }-\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } \right) { x }^{ 2 }+\left( \frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } \right) x-\left( \frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } ,\frac { 1 }{ \gamma } \right) \)
\(\Rightarrow { x }^{ 3 }+\frac { 3 }{ 4 } { x }^{ 2 }+\frac { 1 }{ 2 } x+\frac { 1 }{ 4 } =0\)
Multiplying by 4 we get, 
4x³ + 3x² + 2x + 1 = 0

Here ∝ = 1, β = 1 and ૪ = -2
∴ The required cubic equation is
x³-(1+1-2)x²(1-2-2)x-(1)(1)(-2) = 0
x³- 0x²-3x+2 = 0
x³-3x-2 = 0

Given 2i + 3 is a root
∴ Its conjugate 3 - 2i is also a root of the polynomial equation.
∴ Sum of the roots 3 + 2i + 3 - 2i = 6
Product of the roots = (3 + 2i) (3 - 2i)
= 3² + 2² = 9+ 4 = 13
∴ The polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
⇒x² -x (6) + 13 = 0
⇒ x²- 6x + 13 = 0

Since \(2+i\sqrt { 3 } \) is a root of the polynomial equation, its conjugate 2-i\(\sqrt3\) is also a root of the equation:
∴ Sum of the roots \(=2+i\sqrt { 3 } +2-i\sqrt { 3 } =4\)
Product of the roots \(=(2+i\sqrt { 3 } )(2-i\sqrt { 3 } )\)
\(={ 2 }^{ 2 }+{ (\sqrt { 3 } })^{ 2 }\)
\([\because (a+ib)(a-ib)={ a }^{ 2 }+{ b }^{ 2 }]\)
= 4 + 3 = 7
Hence, the polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x(4)+7=0\)
\(\Rightarrow { x }^{ 2 }-4x+7=0\)

The roots are rational if Δ = b²−4ac = (−2p)²−4(p²−q²+2qr−r²).
But this expression reduces to 4(q²−2qr+r²) or 4(q−r)² which is a perfect square.
Hence the roots are rational.

Since 2-\(\sqrt{3}\)i is a root of the required polynomial equation with real coefficients, 2 +\(\sqrt{3}\)i is also a root. Hence the sum of the roots is 4 and the product of the roots is 7. Thus x²-4x + 7= 0 is the required monic polynomial equation.

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By Vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The required quadratic equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

The roots of x³+2x2+3x+4 = 0 are ∝, β, ૪
∴ ∝+β+૪ = -co-efficient of x² = -2 ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3 ....(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4 ...(3)
Form a cubic equation whose roots are 2∝, 2β, 2૪
2∝+2β+2૪ = 2(∝+β+૪) = 2(-2) = -4 [from (1)]
4∝β+4β૪+4૪∝ = 4(∝β+β૪+୪∝) = 4(3) = 12 [from (2)]
(2∝)(2β)(2૪) = 8(∝β૪) = 8(-4) = -32 [from (3)]
∴ The required cubic equation is
x³-(2∝+2β+2૪)x² + (2∝β+2β૪+2୪∝)x - (2∝)(2β)(2૪) = 0
⇒ x³+(-4)x²+12x+32 = 0
⇒ x³+4x²+12x+32 = 0

Given roots are 1, 2 and 3
Here a = 1, β = 2 and ૪ = 3
A cubic polynomial equation whose roots are α, β, ૪ is
x³-(α+β+૪)+x²(αβ+β૪+૪α)x-∝β૪ = 0
⇒ x³-(1+1+2)x²(2+6+3)x-6 = 0
⇒ x³-6x²+11x-6 = 0

Let the roots be a−d, a, a+d.
Then the sum of the roots is 3a which is equal to 6 from the given equation.
Thus 3a = 6 and hence a = 2.
The product of the roots is a³− ad² which is equal to −24 from the given equation.
Substituting the value of a, we get 8−2d² = −24 and hence d = ±4.
If we take d = 4 we get −2, 2, 6 as roots and if we take d = −4, we get 6, 2, −2 as roots (same roots given in reverse order) of the equation.

Let the roots be in A.P. Then, we can assume them in the form α-d, α, α+d
Applying the Vieta’s formula (α-d)+α+(α+d) = \(\frac{p}{1}\) = p ⇒ 3α = -p ⇒ α = -\(\frac{p}{3}\).
But, we note that α is a root of the given equation. Therefore, we get
\(\left( -\frac { p }{ 3 } \right) ^{ 3 }+p\left(- \frac { p }{ 3 } \right) ^{ 3 }+q\left(- \frac { p }{ 3 } \right) ^{ 3 }+r\) = 0 ⇒ 9 pq = 2p³+27r.

The discriminant Δ =((4p)² - 4(4)(p+2) = 16(p²-p-2) = 16(p+1)(p-2). So we get
Δ < 0 if -1 < p < 2
Δ = 0 if p = -1 or p = 2
Δ >0 if \(\infty\). Thus the given polynomial has
imaginary roots if -1 < p < 2
equal real roots if p = −1 or p = 2;
distinct real roots if -\(\infty\) < p < -1 or 2 < p < \(\infty\)

The cubic equation is
x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0
\( x^{3}-x^{2}\left(2+\frac{1}{2}+1\right)+x\left(1+\frac{1}{2}+2\right)-(2)\left(\frac{1}{2}\right)(1)=0\)
\(x^{3}-x^{2}\left(\frac{4+1+2}{2}\right)+x\left(\frac{2+1+4}{2}\right)-1=0 \)
\(x^{3}-x^{2}\left(\frac{7}{2}\right)+x\left(\frac{7}{2}\right)-1=0\)
2x³ – 7x² + 7x – 2 = 0

Here Δ = b²−4ac = 0 for equal roots. This implies 4(k + 2)² = 4(9)k. This implies k = 4 or 1.

Δ = b²- 4ac = -20 < 0. The roots are imaginary numbers.