Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

Since α and β are the roots of 17x²+ 43x −73 = 0 , we have α + β =\(\frac { -43 }{ 17 } \) and  αβ =\(\frac { -73 }{ 17 } \).
We wish to construct a quadratic equation with roots α + 2 and  β + 2. Thus, to construct such a quadratic equation, calculate
the sum of the roots =  α + β + 4 = \(\frac { -4 }{ 17 } +4=\frac { 25 }{ 17 } \) and
the product of the roots = αβ + 2(α+β)+4 = \(\frac { -73 }{ 17 } +2\left( \frac { -43 }{ 17 } \right) +4=\frac { -91 }{ 17 } \)
Hence a quadratic equation with required roots is x²-\(\frac { 25 }{ 17 } x-\frac { 91 }{ 17 } \) = 0
Multiplying this equation by 17, gives 17x²−25x−91 = 0
which is also a quadratic equation having roots α + 2 and β + 2

If P(x) denotes the polynomial in the equation, then P(2) = 0.
Hence 2 is a root of the polynomial.
To find other roots, we divide the given polynomial x³−5x²−4x + 20 by x − 2 and get Q(x) = x² −3x−10 as the quotient.
Solving Q(x) = 0 we get −2 and 5 as roots.
Thus 2, −2, 5 are the solutions of the given equation.

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

The length and breadth of the cuboid are x + 1.
x + 2 and x + 3
[∵ they are increased by 1, 2, 3 units]
Also volume = V + 52 .
[since V is increased by 52]
∴ V + 52 = (x +1)(x + 2)(x + 3) .........(1)
⇒ V = (x + 1) (x + 2) (x + 3) - 52
Here a = -1, β = -2, ૪ = -3
⇒ V = x³-x²(α+β+૪)+x(αβ+β૪+૪α)-αβ૪ = 52
⇒ V = x³-x²(-1-2-3)+x(2+6+3)-(-1)(-2)(-3) = 52
⇒ V = x³-x²(-6)+x(11)+6-52
⇒ x³ = x²+6x²+11x+6-52
⇒ 6x²+11x-46 = 0
⇒ (6x+23)(x-2) = 0
⇒ (6x+23)(x-2) = 0
⇒ x = 2

∴ Volume of the cube = x³ = 2³ = 8.
Volume of a cuboid = 52 + 8 = 60
[∵ x = \(\frac{-23}{6}\) is not possible as x represents the side of the cube]

According to our notations, a_n= 2 and a₀ = 3.
If \(\frac{p}{q}\) is a zero of the polynomial, then as (p, q) = 1, p must divide 3 and q must divide 2.
Clearly, the possible values of p are 1, −1, 3, −3 and the possible values of q are 1, −1, 2, −2.
Using these p and q we can form only the fractions \(\pm \frac { 1 }{ 1 } ,\pm \frac { 1 }{ 2 } ,\pm \frac { 3 }{ 2 } ,\pm \frac { 3 }{ 1 } \). 
Among these eight possibilities, after verifying by substitution, we get \(\frac { -3 }{ 2 } \) is the only rational zero.
To find other roots, we divide the given polynomial 2x³+ 3x²+ 2x + 3 by 2x + 3 and get x²+1 as the quotient with zero remainder. Solving x²+1 = 0, we get i and −i as roots. Thus \(\frac { -3 }{ 2 } \), -i, i are the roots of the given polynomial equation.

Since the sum of the co-efficients is
2 - 9 + 10 - 3 = 12 - 12 = 0
⇒ x = 1 is a root of (x)
∴ (x - 1) is a factor of (x)
To find the other factor, let us divide f(x) by x-1 

[Using synthetic division]
f(x) = (x-1)(x - 3)(2x - 1) = 0
⇒x - 1 = 0, x - 3 = 0 or 2x - 1 = 0
⇒ x = 1, x = 3x, x = \(\frac{1}{2}\)
Hence the roots are 1, 3, \(\frac{1}{2}\).

Let the roots form a GP be \(\frac{a}{\lambda}\), a, a\(\lambda\)
Product of the roots \(\frac{a}{\lambda} \times a \times a \lambda=\frac{24}{3}=8\)
a³ = 8
a = 2
Sum of the roots, \(\frac{a}{\lambda}+a+a \lambda=\frac{26}{3}\)
\( a\left(\frac{1}{\lambda}+1+\lambda\right) =\frac{26}{3} \)
\(2\left(\frac{1+\lambda+\lambda^{2}}{\lambda}\right) =\frac{26}{3} \)
\( 3+3 \lambda+3 \lambda^{2} =13 \lambda \)
\(3 \lambda^{2}+3 \lambda-13 \lambda+3 =0 \)
\(3 \lambda^{2}-10 \lambda+3 =0\)
\( (\lambda-3)(3 \lambda-1)=0 \)
\( \lambda=3 \text { (or) } \lambda=\frac{1}{3}\)
\( \lambda=3, \quad \frac{a}{\lambda}=\frac{2}{3} , \)
\( a \lambda=2(3)=6 \)
If \( \lambda=\frac{1}{3}, \frac{a}{\lambda}=\frac{2}{\frac{1}{3}}=6 \)
If \( a \lambda=2\left(\frac{1}{3}\right)=\frac{2}{3} \)
Roots are \(\frac{a}{\lambda},\ a,\ a \lambda\)
If \(\lambda\) = 3 roots are \(\frac{2}{3}\), 2, 6
If \(\lambda=\frac{1}{3}\) roots are 6, 2, \(\frac{2}{3}\)

Here, a = 9, b = - 36, c = 44, d = -16
Since the roots form an arithmetic progression,
Let the roots be a - d, a and a + d
Sum of the roots \(=\frac { -b }{ a } \)
\(\Rightarrow (a-d)+(a)+(a+d)=\frac { -(-36) }{ 9 } =4\)
\(\Rightarrow 3a=4\Rightarrow a=\frac { 4 }{ 3 } \)
and product of the roots \(=\frac { -d }{ a } \)
\(=\frac { -(-16) }{ 9 } =\frac { 16 }{ 9 } \)
\(\Rightarrow (a-d)(a)(a+d)=\frac { 16 }{ 9 } \)
\(\left( { a }^{ 2 }-{ d }^{ 2 } \right) (a)=\frac { 16 }{ 9 } \)
\(\left( \frac { 16 }{ 9 } -{ d }^{ 2 } \right) \left( \frac { 4 }{ 3 } \right) =\frac { 16 }{ 9 } \ [\because a=\frac { 4 }{ 3 } ]\)
\(\frac { 16 }{ 9 } -{ d }^{ 2 }=\frac { 16 }{ 9 } \times \frac { 3 }{ 4 } =\frac { 4 }{ 3 } \)
\(\frac { 16 }{ 9 } -\frac { 4 }{ 3 } ={ d }^{ 2 }=\frac { 16 }{ 9 } \times \frac { 3 }{ 4 } =\frac { 4 }{ 3 } \)
\(\frac { 16 }{ 9 } -\frac { 4 }{ 3 } ={ d }^{ 2 }\)
\(\Rightarrow { d }^{ 2 }=\frac { 16-12 }{ 9 } =\frac { 4 }{ 9 } \)
\(\Rightarrow d=\pm \sqrt { \frac { 4 }{ 9 } } =\frac { 2 }{ 3 } \)
∴ The roots are a-d, a, a+d
\(\Rightarrow \frac { 4 }{ 3 } -\frac { 2 }{ 3 } ,\frac { 4 }{ 3 } ,\frac { 4 }{ 3 } +\frac { 2 }{ 3 } \Rightarrow \frac { 2 }{ 3 } ,\frac { 4 }{ 3 } ,2\)

Given equation is 2x⁴- 8x + 6x²- 3 = 0
Here a = 2, b = -8, c = 6, d = 0, e = -3
Let ∝, β, ૪ and \(\delta \) be the roots of equation (1)
Then by Vieta's formula,
\(\sum { _{ 1 }= } \alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -(-8) }{ 2 } =4\)
\(\sum { _{ 2 } } =\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { 6 }{ 2 } =3\)
\(\sum { _{ 3 } } =\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =\frac { 0 }{ a } \)
\(\sum { _{ 4 } } =\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { -3 }{ 2 } \)
Now, (a+b+c+d)² = a²+b²+c²+d²+2(ab+ac+ad+bc+cd)
⇒ n∝²+β²+૪²+\(\delta\)² = (∝ + β + ૪ + \(\delta\))²-2(\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta \))
∝² + β² + ૪² = 4²-2(3) = 16 - 6 = 10