Given, \(\alpha+\beta=24\)            ... (i)
\(\alpha-\beta=8\)                          ... (ii)
Adding equations (i) and (ii),
\(2\alpha=32\)
\(\Rightarrow \quad \alpha=16\)
Put the value of \(\alpha\) in equation (i),
\(16+\beta=24\)
\(\Rightarrow \quad \beta=24-16 =8\)
Hence, the quadratic polynomial = x² - (Sum of zeroes)x + (Product of zeroes)
\(={ x }^{ 2 }-\left( \alpha +\beta \right) x+\alpha \beta \)
\(={ x }^{ 2 }-\left( 16+8 \right) x+(16)(8)\)
\(={ x }^{ 2 }-24 x+128\)
Verification : \(\alpha+\beta=\frac{-b}{a}=-\frac { Coeff. \ of \ x }{ Coeff. \ of \ { x }^{ 2 } } \)
\(\Rightarrow \quad 24=-\left( -\frac { 24 }{ 1 } \right) \)
and \(\alpha \beta =\frac { c }{ a } =\frac { Costant \ term }{ Coeff. \ of \ { x }^{ 2 } } \)
\(\Rightarrow \quad 128=\left( \frac { 128 }{ 1 } \right) \)
Hence, the relationship is verified. 

Let  p(x) = 5x² + 8x - 4
= 5x² + 10x - 2x - 4
= 5x (x + 2) - 2 (x + 2)
= (x + 2) (5x - 2)
Hence, zeroes of the quadratic polynomial 5x² + 8x - 4 are - 2 and \(\frac{2}{5}\)
Verification :
Sum of zeroes = \(-2+\frac{2}{5}=\frac{-8}{5}\)
Product of zeroes = \((-2)\times(\frac{2}{5})=\frac{-4}{5}\)
Again sum of zeroes  \(=-\frac { Coeff. \ of \ x }{ Coeff. \ of \ { x }^{ 2 } } =\frac { -8 }{ 5 } \)
Product of zeroes = \(=\frac { Costant \ term }{ Coeff. \ of \ { x }^{ 2 } } =\frac { -4 }{ 5 } \)
Thus relationship is verified.

f(x) = 4x² + 4x - 3
\(\Rightarrow \quad f\left( \frac { 1 }{ 2 } \right) =4\left( \frac { 1 }{ 4 } \right) +4\left( \frac { 1 }{ 2 } \right) -3\)
= 1 + 2 - 3 = 0
and   \(f\left( \frac { -3 }{ 2 } \right) =4\left( \frac { 9 }{ 4 } \right) +4\left( -\frac { 3 }{ 2 } \right) -3\)
= 9 - 6 - 3 = 0
\(\therefore \quad \frac{1}{2},-\frac{3}{2}\) are zeroes of polynomial 4x² + 4x - 3
Sum of zeroes = \(\frac{1}{2}-\frac{3}{2}=-1=\frac{-4}{4}\)
\(=-\frac { Coefficient \ of \ x }{ Coefficient \ of \ { x }^{ 2 } } \)
Product of zeroes = \(\left( \frac { 1 }{ 2 } \right) \left( -\frac { 3 }{ 2 } \right) =\frac { -3 }{ 4 } \)
\(=\frac { Costant \ term }{ Coefficient \ of \ { x }^{ 2 } } \)
\(\therefore\)  Relation between zeroes and coeff.of polynomial is verified.

If \(\alpha\) and \(\beta\) are the zeroes of 2x² - 3x + 1,
then \(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow \quad \alpha+\beta=\frac{3}{2}\)
and \(\alpha\beta=\frac{c}{a}\)
\(\Rightarrow \quad \alpha\beta=\frac{1}{2}\)
New quadratic polynomial whose zeroes are \(3\alpha\) and \(3\beta\) is :
x² - (Sum of the roots)x + Product of the roots
= x² - \((3\alpha+3\beta)x+3\alpha\times3\beta\)
= x² - 3(\(\alpha+\beta\))x + 9\(\alpha\beta\)
\(={ x }^{ 2 }-3\left( \frac { 3 }{ 2 } \right) x+9\left( \frac { 1 }{ 2 } \right) \)
\(={ x }^{ 2 }-\frac { 9 }{ 2 } x+\frac { 9 }{ 2 } \)
\(=\frac { 1 }{ 2 } \left( 2{ x }^{ 2 }-9x+9 \right) \)
Hence, required quadratic polynomial is \(​​\frac { 1 }{ 2 } \left( 2{ x }^{ 2 }-9x+9 \right) \)

(i) Now p(x) = 2x³ - 11x² + 17x - 6
p(2) = 2(2)³ - 11(2)² + 17(2) - 6
= 16 - 44 + 34 - 6
= 50 - 50
= 0
Hence 2 is the zero of p(x)
(ii) Again  p(3) = 2(3)³ - 11(3)² + 17(3) - 6
= 54 - 99 + 51 - 6
= 105 - 105 = 0
Hence, 3 is zero of p(x)
(iii) Again  \(p\left( \frac { 1 }{ 2 } \right) =2\left( \frac { 1 }{ 2 } \right) ^{ 3 }-11\left( \frac { 1 }{ 2 } \right) ^{ 2 }+17\left( \frac { 1 }{ 2 } \right) -6\)
\(=\frac { 1 }{ 4 } -\frac { 11 }{ 4 } +\frac { 17 }{ 2 } -6\)
= 0
Hence, \(\frac{1}{2}\) is also the zero of p(x).

Let P(x)=x²-5
For finding a zeroes of p(x), p(x)=0
x²-5=0
⇒ x=±5
Let α=5 and β=-5
Now, 1+a=1+5=6
and 1-a=1-5=4
Thus, 6 and -4 are the zeroes of new quadratic polynomial.
Therefore the new quadratic polynomial will be (x-6)(x+4) or x²-2x-24

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.

Since, a and β are the zeroes of polynomial x² -px-p+c.
So, sum of zeroes, a+β =p      ...(i)
and product of zeroes, aβ =c-p|   ...(ii)
Also, (a+1)(β+1)=0  [given]
aβ+(a+β)+1=0
⇒ c-p+p+1=0 [from Eqs. (i) and (ii)]
⇒ c-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

Let given polynomial be
p(x)=x⁴ -6x³-26x²+138x-35
Give, \(2\pm \sqrt { 3 } \) are two zeroes of the polynomial.
So, \((x-(2+\sqrt { 3 }) )\) and \((x-(2-\sqrt { 3 }) )\) are factors of p(x)
⇒ \((x-(2+\sqrt { 3 }) )\) \((x-(2-\sqrt { 3 }) )\) is a factor of (p(x)
\(\Rightarrow \quad ((x-2)-\sqrt { 3 } )((x-2)+\sqrt { 3 } )\) is a factor p(x)
\(\Rightarrow \quad (x-2)^{ 2 }-(\sqrt { 3 } )^{ 2 }\) is a factor p(x)
⇒ x²-4x+1 is a factor p(x)
Now, divide p(x) by x²-4x+1 to obtain other zeroes.
Then, the division process is
∴ p(x)=x⁴ -6x³-26x²+138x-35
-(x²-4x+1)(x²-4x+1)(x²-2x-35)
Now, for other zeores put x²-2x-35=0
⇒ x²-7x+5x-35=0               [ by splitting the middle term]
⇒ [x(x-7)+5(x-7)]=0
⇒ (x+5)(x-7)=0⇒x=5, 7
Hence, -5 and 7 are other zeores of the given polynomial.

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.

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

If x + 7 is a factor then (-7) is a root.
So, f(-7) = (-7)⁴ + 10(-7)³ + 25(-7)² + 15(-7) + k = 0
(when it is a root the polynomial should be equal to zero when value is substituted)
2401 - 3430 + 1225 - 105 + k = 0
\(\Rightarrow\)  3626 - 3535 + k = 0
\(\Rightarrow\)  91 + k = 0
\(\therefore\)  k = - 91

Sum of zeroes \(=-\frac { Coefficient\ of \ x }{ Coefficient \ of \ { x }^{ 2 } } \) 
\(\therefore \quad a + b = - a\)
\(\Rightarrow \quad 2a + b = 0\)
Product of zeroes = \(​​=-\frac { Constant \ term }{ Coefficient \ of \ { x }^{ 2 } } \)
\(\therefore \quad ab = b\)
\(\Rightarrow \quad a=1\)
then b = - 2

\({ x }^{ 2 }-4\sqrt { 3 } x+3=0\)
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 \(\alpha\beta=\frac{c}{a}\)
\(\Rightarrow \quad \alpha\beta=\frac{3}{1}\)
\(\Rightarrow \quad \alpha\beta=3\)
\(\therefore \quad \alpha +\beta -\alpha \beta =4\sqrt { 3 } -3\)

Let p(x) = 3x² + 11x - 4
= 3x² + 12x - x - 4
= 3x(x + 4) - 1(x + 4)
= (3x - 1)(x + 4)
So, zeroes are \(m=\frac{1}{3}\) and n = - 4
Now, \(\frac { m }{ n } +\frac { n }{ m } =\frac { \left( \frac { 1 }{ 3 } \right) }{ -4 } +\frac { -4 }{ \left( \frac { 1 }{ 3 } \right) } =\frac { 1 }{ -12 } -12\)
\(=\frac{-145}{12}\)

Comparing x²+4x+2a with ax²+bx+c, 
a=1,b=4,c=2a
According to the question,
\(\alpha, \frac{\alpha}{2}\) are the zeroes of the polynomial,
sum of the zeroes \(\frac{\mathrm{b}}{\mathrm{a}}\) 
\(\alpha+\frac{\alpha}{2}=\frac{-4}{1}\) 
\(\alpha=\frac{-8}{3}\) 
\(\alpha^{2}=\frac{64}{9}\) [squaring both sides]
product of all zeroes,
\(\alpha \times \frac{\alpha}{2}=\frac{2 a}{1}\) 
α²=4a
\(\therefore 4 \mathrm{a}=\frac{64}{9}\left[\because \alpha^{2}=\frac{64}{9}\right]\) 
\(a=\frac{16}{9}\)

f(x) = x² - 7x - 8
Let other zero be k, then
Sum of zeroes \(-1+k=-\left( \frac { -7 }{ 1 } \right) =7\)
\(\Rightarrow k = 8\)

p(x) = 3x² - kx + 6
Sum of the zeroes = 3 \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad 3=-\frac { \left( -k \right) }{ 3 } \)
k = 9

Sum of the roots \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad \alpha +\beta =-\left( \frac { b }{ a } \right) \)
\(=\frac{b}{a}\)

\(-\frac { 145 }{ 12 } \)

k=7