p(x) = x² + 7x + 9
q(x) = x + 2
r(x) = - 1
g(x) = ?
p(x) = g(x) q(x) + r(x)
\(\Rightarrow\)  x² + 7x + 9 = g(x)(x + 2) - 1
\(\Rightarrow \quad g\left( x \right) =\frac { { x }^{ 2 }+7x+10 }{ x+2 } \)
\(\frac { \left( x+2 \right) \left( x+5 \right) }{ x+2 } \)
\(\therefore \quad g(x) =x+5\)

p(y) = 6y² - 7y + 2
\(\alpha +\beta =-\left( -\frac { 7 }{ 6 } \right) =\frac { 7 }{ 6 } \)
and \(\alpha\beta=\frac{2}{6}=\frac{1}{3}\)
Now \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { { 7 }/{ 6 } }{ { 2 }/{ 6 } } =\frac { 7 }{ 2 } \)
and \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =3\)
The required polynomial is \({ y }^{ 2 }-\frac { 7 }{ 2 } y+3=\frac { 1 }{ 2 } \left[ 2{ y }^{ 2 }-7y+6 \right] \)

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

x²-15x+54

Let p(x)=2x²+2ax+5x+10
Since, x+a is a factor of p(x).
Therefore, p(-a)=0
⇒ 2(-a)²+2a(-a)+5(-a)+10=0
⇒ 2a²-2a²-5a+10=0⇒5a=10a=2

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
\(\frac{1}{2}-\beta=\frac{1}{6}\)
\(\Rightarrow \quad \frac{1}{2}-\frac{1}{6}=\beta\)
\(\Rightarrow \quad \frac{2}{6}=\frac{1}{3}=\beta\)
\(\therefore \quad \alpha\beta=\frac{k}{6}=\frac{1}{2}\times\frac{1}{3}\)
\(\therefore \quad k=1\)

2, 2

(i) 5
(ii) 3

The given quadratic polynomial is f(x) = x² + px + 45. Let \(\alpha\) and \(\beta\) be the zeroes of the given quadratic polynomial.
\(\therefore \quad \alpha +\beta =-p\) and \(\alpha\beta=45\)           ....(i)
Given, \(\left( \alpha -\beta \right) ^{ 2 }=144\)
\(\Rightarrow \quad \left( \alpha +\beta \right) ^{ 2 }-4\alpha \beta =144\)
\(\Rightarrow \quad \left( -p \right) ^{ 2 }-4\times 45=144\)
\(\Rightarrow \quad p^{ 2 }-180=144\)
\(\Rightarrow \quad p^{ 2 }=144+180=324\)
\(\therefore \quad p=\pm \sqrt { 324 } =\pm 18\)
Thus, the value of p is \(\pm 18\)

Let f(x) = 2x² - 5x - 3
Let the zeroes of polynomial be \(\alpha\) and \(\beta\), then
Sum of zeroes = \(\alpha+\beta = \frac{5}{2}\)
Product of zeroes = \(\alpha\beta=-\frac{3}{2}\)
According to the question, zeroes of x² + px + q are \(2\alpha\) and \(2\beta\)
Sum of zeroes \(-\frac { Coeff.of \ x }{ Coeff.of \ { x }^{ 2 } } =\frac { -p }{ 1 } \)
\(\Rightarrow \quad -p= 2\alpha+2\beta=2(\alpha+\beta)\)
\(\Rightarrow \quad -p=2\times\frac{5}{2}=5\Rightarrow p=-5\)
Product of zeroes \(=\frac { Constant \ term }{ Coeff \ of \ { x }^{ 2 } } =\frac { q }{ 1 } \)
\(\Rightarrow \quad q = 2\alpha\times2\beta=4\alpha\beta\)
\(\Rightarrow \quad q=4(-\frac{3}{2})=-6\)
\(\therefore \quad p=-5\) and \(q=-6\)

f(x) = 2x² - 7x + 3
Sum of roots = p + q \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(=-\left( \frac { -7 }{ 2 } \right) =\frac { 7 }{ 2 } \)
Product of roots = pq \(=-\frac { Constant\quad term }{ Coefficient\quad of\quad { x }^{ 2 } } =\frac{3}{2}\)
We know that 
(p + q)² = p² + q² + 2pq
\(\Rightarrow\)  p² + q² = (p + q)² - 2pq
\(=\left( \frac { 7 }{ 2 } \right) ^{ 2 }-3=\frac { 49 }{ 4 } -\frac { 3 }{ 1 } =\frac { 37 }{ 4 } \)

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

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,
\(\alpha \times \frac { 1 }{ \alpha } =\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } \) 
\(\\ \Rightarrow \ 1=\frac { 6\alpha }{ a^{ 2 }+9 }\) 
\( \\ \Rightarrow \ a^{ 2 }+9=6a\) 
\(\\ \Rightarrow \ a^{ 2 }-6a+9=0\) 
\(\\ \Rightarrow \ (a-3)^{ 2 }=0\ \ \left[ \because \quad (x-y)^{ 2 }=x^{ 2 }+y^{ 2 }-2xy \right] \) 
\(\\ \therefore \ a=3\)

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

Quotient, q(x) = x² - 3x + 2 and remainder, r(x) = 2.

 -2,3