From the given polynomial we will find the value, the sum of the zeroes, and the multiple of the zeroes.
\(\alpha+\beta=\frac{-b}{a}=\frac{-8}{1}=-8\)
\(\alpha\times\beta=\frac{c}{a}=\frac{6}{1}=6\)
Sum of zeroes = \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { -8 }{ 6 } =\frac { -4 }{ 3 } \)
Product of zeroes = \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ 6 } \)
Now for making a polynomial 
p(x) = x² - \((\alpha+\beta)x+\alpha\beta\)
\(\therefore\)  The polynomial is : \(p(x)=\frac { 1 }{ 6 } \left( 6{ x }^{ 2 }+8x+1 \right) \)

Given, \(\alpha\) and \(\frac{2}{\alpha}\) are the zeroes of x² + 4x + 2a.
We know that,
Product of the zeroes = \(=\frac { Constant\quad term }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad \alpha\times\frac{2}{\alpha}=\frac{2a}{1}\)
\(\Rightarrow 2=2a\)
\(\therefore \quad a=1\)

Here, the graph of p(x) intersects X-axis at one point only 
\(\therefore\) Number of zeroes of p(x) = 1

Let \(f(y)=7y^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 }\)
 \(=\frac { 21y^{ 2 }-11y-2 }{ 3 } \)
\(=\frac { 21y^{ 2 }-14y+3y-2 }{ 3 } \)  [by splitting the middle term]
\(=\frac { 7y(3y-2)+1(3y-2) }{ 3 } \)
\(=\frac { 1 }{ 3 } (3y-2)(7y+1)\)
So, the value of \(7y^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 } \) is zero whe 3y-2=0 or 7y+1=0, i.e. when \(y=\frac { 2 }{ 3 } \) or \(y=\frac { 1 }{ 7 } \).
Thus, the zeroes\(=\frac { 2 }{ 3 } -\frac { 1 }{ 7 } =\frac { 14-3 }{ 21 } =\frac { 11 }{ 21 } =-\left( \frac { -11 }{ 3\times 7 } \right) \)
\(=-(1).\left( \frac { Coefficient \ of \ y }{ Coefficient \ of \ y^{ 2 } } \right) \) and product of zeroes\(=\left( \frac { 2 }{ 3 } \right) \left( -\frac { 1 }{ 7 } \right) =\frac { -2 }{ 21 } =\frac { -2 }{ 3\times 7 } \)
\(=\left( \frac { Constant \ term }{ Coefficient \ of \ y^{ 2 } } \right) \)
Hence, the relations between the zeroes and the coefficients of the polynomials is verified.

Given, a and β are the zeroes of f(x)=ax²+bx+c.
\(\alpha +\beta =-\frac { Coefficient \ of \ x }{ Coefficient \ of\ x^{ 2 } } =-\frac { b }{ a } \)    ....(i)
and \(\alpha \beta =\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } =\frac { c }{ a } \) .....(ii)
We know that,
\(\left( \alpha -\beta \right) ^{ 2 }=\left( \alpha +\beta \right) ^{ 2 }-4\alpha \beta \)
\(\left( \alpha -\beta \right) ^{ 2 }=\left( \frac { -b }{ a } \right) ^{ 2 }-4\left( \frac { c }{ a } \right) \)   [from Eqs. (i) and (ii)]
\(=\frac { b^{ 2 } }{ a^{ 2 } } -\frac { 4c }{ a } =\frac { b^{ 2 }-4c }{ a^{ 2 } } \)
On taking square root both sides, we get
\(\alpha -\beta =\pm \frac { 1 }{ a } \sqrt { b^{ 2 }-4c } \)
which is the required difference between their zeroes.

Given polynomial is p(x)=4x²+2x-5a.
Since, 2 is a zero of polynomial.
p(2)=0
⇒ 4(2)²+2(2)-5a=0     [putting x=2]
⇒ 16+4-5a=0 ⇒ 5a=20
\(\therefore\) \(a=\frac { 20 }{ 5 } =4\)

In the polynomial \(7q^{ 6 }+4q^{ 2 }+\frac { 3 }{ 2 } +q-8\) , the highest power of the variable q is 6.

\({ x }^{ 2 }-4\sqrt { 3 } x+3=0\)
If \(\alpha\) and \(\beta\) are the zeroes of \({ x }^{ 2 }-4\sqrt { 3 } x+3\)
then \(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow \quad \alpha +\beta =-\frac { \left( -4\sqrt { 3 } \right) }{ 1 } \)
\(\Rightarrow \quad \alpha +\beta=4\sqrt { 3 }\)
and \(\alpha\beta=\frac{c}{a}\)
\(\Rightarrow \quad \alpha\beta=\frac{3}{1}\)
\(\Rightarrow \quad \alpha\beta=3\)
\(\therefore \quad \alpha +\beta -\alpha \beta =4\sqrt { 3 } -3\)

p(x) = ax² + bx + c
Let \(\alpha\) and \(\frac{1}\alpha\) be the zeroes of p(x),  then
Product of zeroes, \(\alpha\times\frac{1}\alpha=\frac{c}{a}\)
So,required condition is, c = a.

According to the question,
Sum of zeroes = \(\frac{21}{8}\)
and Product of zeroes = \(\frac{5}{16}\)
So, quadratic polynomial = x² - (Sum of zeroes)x + Product of zeroes
= x² - \(\left( \frac { 21 }{ 8 } \right) \)x + \(\left( \frac { 5 }{ 16 } \right) \)
= \(\frac { 1 }{ 16}\left( 16{ x }^{ 2 }-42x+5 \right) \)
 = \(\left( 16{ x }^{ 2 }-42x+5 \right) \frac { 1 }{ 16}\)

f(x) = x² - 2x
= x(x - 2)
i.e. f(x) = 0 \(\Rightarrow \)  x = 0 or x = 2
Hence zeroes are 0 & 2.

x²-x-56

Let α and \(\frac { 1 }{ \alpha } \) be two zeroes of the given polynomial, which are reciprocal of each other.
On comparing the given polynomial with Ax²+Bx+C, we get
A=a²+9, B=13 and C=6a
Now, product of zeroes,
\(\alpha \times \frac { 1 }{ \alpha } =\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } \) 
\(\\ \Rightarrow \ 1=\frac { 6\alpha }{ a^{ 2 }+9 }\) 
\( \\ \Rightarrow \ a^{ 2 }+9=6a\) 
\(\\ \Rightarrow \ a^{ 2 }-6a+9=0\) 
\(\\ \Rightarrow \ (a-3)^{ 2 }=0\ \ \left[ \because \quad (x-y)^{ 2 }=x^{ 2 }+y^{ 2 }-2xy \right] \) 
\(\\ \therefore \ a=3\)

We know that a polynomial in one variable x, is an algebraic expression of the form,
\(p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}\)  where n is a positive integer and a₀, a₁, a₂,..., a_n are constants.
(i) No, because powers of any polynomial is always positive integer.
(ii) Yes, because it satisfies the definition of polynomial.
(iii) No, because powers of any polynomial can not be negative integer.
(iv) Yes, because it satisfies the condition of polynomial.

\(a=-\frac { 3 }{ 2 } \)

 Quadratic

Since \(\alpha\) and \(\beta\) are the zeroes of the cubic polynomial x²  +4x + 3
then, \(\alpha+\beta=-4 \)
and \(\alpha\beta=3\)
Sum of zeroes  \(=1+\frac { \beta }{ \alpha } +1+\frac { \alpha }{ \beta } \)
\(=\frac { \alpha \beta +{ \beta }^{ 2 }+\alpha \beta +{ \alpha }^{ 2 } }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta }{ \alpha \beta } \)
\(=\frac { \left( \alpha +\beta \right) ^{ 2 } }{ \alpha \beta } =\frac { \left( -4 \right) ^{ 2 } }{ 3 } =\frac { 16 }{ 3 } \)
Product of zeroes = \(\left( 1+\frac { \beta }{ \alpha } \right) \left( 1+\frac { \alpha }{ \beta } \right) \)
\(=1+\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha } +\frac { \alpha \beta }{ \alpha \beta } \)
\(\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta }{ \alpha \beta } =\frac { \left( \alpha +\beta \right) ^{ 2 } }{ \alpha \beta } \)
\(=\frac { \left( -4 \right) ^{ 2 } }{ 3 } =\frac { 16 }{ 3 } \)
But required polynomial = x² - (Sum of the zeroes) x + Product of the zeroes
\(={ x }^{ 2 }-\left( \frac { 16 }{ 3 } \right) x+\frac { 16 }{ 3 } \)
or \(=\left( { x }^{ 2 }-\frac { 16 }{ 3 } x+\frac { 16 }{ 3 } \right) \)
\(=\left( 3{ x }^{ 2 }-16x+16 \right) \frac { 1 }{ 3 } \)

Factors of x² + 7x + 12 :
x² + 7x + 12 = 0
\(\Rightarrow\)  x² + 4x + 3x + 12 = 0
\(\Rightarrow\)  x(x + 4) + 3 (x + 4) = 0
\(\Rightarrow\)  (x + 4) (x + 3) = 0
\(\Rightarrow\)  x = - 4, - 3                .... (i)
Let p^'(x) = x⁴ + 7x³ + 7x² + px + q
If p(x) is exactly divisible by x² + 7x + 12, then x = - 4 and x = - 3 are zeroes of p(x)  [from eq (i)]
p(x) = x⁴ + 7x³ + 7x² + px + q
p(- 4) = (-4)⁴ + 7(-4)³ + 7(-4)² + p(-4) + q
but p(-4) = 0
\(\therefore\)  0 = 256 - 448 + 112 - 4p + q
\(\Rightarrow\)  0 = - 4p + q - 80
\(\Rightarrow\)  4p - q = 80                       ... (ii)
and  p(-3) = (-3)⁴ + 7 (-3)³ + 7(-3)² + p(-3) + q
but p(-3) = 0
\(\therefore\)   0 = 81 - 189 + 63 - 3p + q
\(\Rightarrow\)  0 = -3 p+ q - 45
\(\Rightarrow\)  3p - q = - 45
On solving eq.(ii) and eq. (iii) by elimination method, we get
4p - q = - 80
3p - q = - 45
p = - 35
On putting the value of p in eq. (i),
4(- 35) - q = - 80
\(\Rightarrow\)  - 140 - q = - 80
\(\Rightarrow\)  - q = 140 - 80
\(\Rightarrow\)  - q = 60
\(\Rightarrow\)   q = - 60
Hence, p = - 35, q = - 60

Let p(x)=2x⁴-3x³-3x²+6x-2.
Since, its two zeroes are \(\sqrt { 2 } \) and \(-\sqrt { 2 } \).
So, (x-\(\sqrt { 2 } \)) and (x+\(-\sqrt { 2 } \)) are the factors of p(x).
⇒  (x-\(\sqrt { 2 } \)) (x+\(-\sqrt { 2 } \)) is a factor of p(x).
⇒ x²-2 also a factor of p(x).
Now, let us divide the given polynomial p(x) by g(x)=x²-2. Then division process is
Here, quotient=2x²-3x+1 and remainder=0
Now, factorise the quotient by splitting the middle term, i.e. write
2x²-3x+1=2x²-2x-x+1
=2x(x-1)-1(x-1)=(2x-1)(x-1)
So, other zeroes of f(x) are given by
(2x-1)=0 and (x-1)=0
⇒ \(x=\frac { 1 }{ 2 } \) and x=1
Hence, all the zeroes of  2x⁴-3x³-3x²+6x-2 are \(\sqrt { 2 } \), \(-\sqrt { 2 } \), \(\frac{1}{2}\) and 1.

Given, α and β are the zeroes of the quadratic polynomial p(s)=3s²-6s+4.
\(\therefore \ a+\beta =-\frac { Coefficient \ of \ s }{ Coefficient \ of \ s^{ 2 } } =\frac { -(-6) }{ 3 } =\frac { 6 }{ 3 } =2\)
and \(a\beta =\frac { Constant \ term }{ Coefficient \ of \ s^{ 2 } } =\frac { 4 }{ 3 } \)
Now, \(\frac { a }{ \beta } +\frac { \beta }{ a } +2\left( \frac { 1 }{ a } +\frac { 1 }{ \beta } \right) +3a\beta \)
\(=\frac { a^{ 2 }+\beta ^{ 2 } }{ a\beta } +2\left( \frac { a+\beta }{ a\beta } \right) +3a\beta \)
\(=\frac { (a+\beta )^{ 2 } }{ a\beta } +2\left( \frac { a+\beta }{ a\beta } \right) +3a\beta \)
\(\left[ \because \quad a^{ 2 }+b^{ 2 }=(a+b)^{ 2 }-2ab \right] \)
\(=\frac { (2)^{ 2 }-2(4/3) }{ 4/3 } +2\left( \frac { 2 }{ 4/3 } \right) +3\times \frac { 4 }{ 3 } \)
\(\quad \left[ \because \quad a+\beta =2\quad and\quad a\beta =4/3 \right] \)
\(=\frac { 4-\frac { 8 }{ 3 } }{ \frac { 4 }{ 3 } } +2\times 2\times \frac { 3 }{ 4 } +3+4\)
\(=\frac { 4 }{ 3 } \times \frac { 3 }{ 4 } +7=1+7=8\)

(i) On dividing p(x) by g(x), we get
Quotient, q(x)=x+2 and remainder, r(x)=(k-21)x
Since, p(x) is exactly divisible by g(x), so remainder =0
⇒ (k-21) x=0⇒ (k-21) x=0.x
On comparing coefficient of x, we get
k-21=0⇒k=21

\(\frac { \alpha }{ \beta } +\frac { \alpha }{ \alpha } +2\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } \right) +3\alpha \beta \)
\(=\frac { \alpha ^{ 2 }+\beta ^{ 2 } }{ \alpha \beta } +2\left( \frac { \alpha +\beta }{ \alpha \beta } \right) +3\alpha \beta =-\frac { 2 }{ 3 } \)