According to the question,  \(\alpha\) and \(\beta\) are zeroes of the polynomial p(x) = 6x² - 5x + k
So, Sum of zeroes = \(\alpha +\beta =-\left( \frac { -5 }{ 6 } \right) =\frac { 5 }{ 6 } \)            ... (i)
Product of zeroes = \(\alpha \beta =\frac { k }{ 6 } \)
\(\alpha-\beta=\frac{1}{6}\)           (Given)     ...... (ii)
Adding equations (i) and (ii), we get
\(2\alpha=2\)
\(\Rightarrow \quad \alpha=\frac{1}{2}\)
Putting the value of \(\alpha\) in equation (ii), we get
\(\frac{1}{2}-\beta=\frac{1}{6}\)
\(\Rightarrow \quad \frac{1}{2}-\frac{1}{6}=\beta\)
\(\Rightarrow \quad \frac{2}{6}=\frac{1}{3}=\beta\)
\(\therefore \quad \alpha\beta=\frac{k}{6}=\frac{1}{2}\times\frac{1}{3}\)
\(\therefore \quad k=1\)

Since \(\alpha\) and \(\beta\) are the zeroes of polynomial p(x) = 3x² + 2x + 1
Hence, \(\alpha+\beta=-\frac{2}{3}\)
and \(\alpha\beta=\frac{1}{3}\)
Now for the new polynomial,
Sum of the zeroes = \(\frac { 1-\alpha }{ 1+\alpha } +\frac { 1-\beta}{ 1+\beta} \)
\(=\frac { \left( 1-\alpha +\beta -\alpha \beta \right) +\left( 1+\alpha -\beta -\alpha \beta \right) }{ \left( 1+\alpha \right) \left( 1+\beta \right) } \)
\(=\frac { 2-2\alpha \beta }{ \left( 1+\alpha +\beta +\alpha \beta \right) } =\frac { 2-\frac { 2 }{ 3 } }{ 1-\frac { 2 }{ 3 } +\frac { 1 }{ 3 } } \)
\(\therefore\)  Sum of zeroes \(=\frac { { 4 }/{ 3 } }{ { 2 }/{ 3 } } =2\)
Product of zeroes \(=\left[ \frac { 1-\alpha }{ 1+\alpha } \right] \left[ \frac { 1-\beta }{ 1+\beta } \right] \)
\(=\frac { \left( 1-\alpha \right) \left( 1-\beta \right) }{ \left( 1+\alpha \right) \left( 1+\beta \right) } \)
\(=\frac { 1-\alpha -\beta +\alpha \beta }{ 1+\alpha +\beta +\alpha \beta } =\frac { 1-\left( \alpha +\beta \right) +\alpha \beta }{ 1+\left( \alpha +\beta \right) +\alpha \beta } \)
\(\therefore\)  Product of zeroes \(=\frac { 1+\frac { 2 }{ 3 } +\frac { 1 }{ 3 } }{ 1-\frac { 2 }{ 3 } +\frac { 1 }{ 3 } } =\frac { \frac { 6 }{ 3 } }{ \frac { 2 }{ 3 } } =3\)
Hence, Required polynomial = x²  - (Sum of zeroes)x + Product of zeroes
= x² - 2x + 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

The graph of y = p(x) does not intersect the X-axis at any point. So, it has no zero.

Quotient=2x²+5, remainder=x+2

\(1,\frac { 1 }{ 2 } \)

Let g(x)=k, then 2x ³ +5x ² -14x+10+k will be exactly by 2x-3.
On dividing 2x ³ +5x ² -14x+10+k by 2x-3, we get Quotient x²+4x-1 and remainder 7+k.
Since, 2x ³ +5x ² -14x+10+k is exactly divisible by 2x-3, so remainder=0
⇒ 7+k=0⇒k=-7

\(\sqrt { 5 } ,-\sqrt { 5 } \)

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

Let f(x)=Lx⁴+Mx³+Nx²+Rx+P.
Then, we have f(1)=0
and f(-1)=0
⇒ L+M+N+R+P=0    ..(i)
and L-M+N-R+P=0  ...(ii)
From Eq. (ii), we have L+N+P=M+R

Now, for finding zeroes, put x²-8x+12=0
⇒ (x-6)(x-2)=0⇒ x=2, 6
Hence, the required quadratic polynomial is x²-8x+12 and their zeroes are 2 and 6.

Let p(y)=y²+92y+1920=(y+32)(y+60)
Now, for zeroes of p(y), put p(y)=0
Zeroes y=-32, -60

(i) 5
(ii) 3

α+β+αβ=\(-\frac { 7 }{ 2 } +\frac { 5 }{ 2 } =-1\)

\(f(2)=0\Rightarrow \alpha =\frac { 5 }{ 2 } \)