6x² - 3 - 7x = 6x² - 7x - 3
=6x² -9x + 2x -3
=3x(2x - 3) + 1(2x - 3)
=(2x - 3)(3x + 1)
\(=2\left(x-\frac{3}{2}\right) 3\left(x+\frac{1}{2}\right)\) 
\(=6\left(x-\frac{3}{2}\right)\left(x+\frac{1}{3}\right)\) 
The zeroes of the polynomial are {3/2, -1/3}
Relationship between the zeroes and the coeffieicients of the polynomial
Also sum of the zeroes = \(\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}=\frac{7}{6}\) 
Also product of the zeroes = \(\frac{3}{2} \times-\frac{1}{3}=-\frac{1}{2}\) 
Hence Verified.

4s² – 4s + 1 = 4s² – 2s - 2s + 1
=2s(2s - 1) - 1(2s - 1)
=(2s - 1)(2s -1)
\(=2\left(s-\frac{1}{2}\right) 2\left(s-\frac{1}{2}\right)\) 
Therefore the zeroes of the polynomial are 1/2, 1/2 
Relationship between the zeroes and the coefficient of the polynomial: 
Also sum of the zeroes = \(\frac{1}{2}+\frac{1}{2}=1\) 
Also product of the zeroes = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\) 
hence verfied.

Let p(x) = x² - 2x - 8
= x² - 4x + 2x - 8 [by splitting the middle term]
= x(x - 4) + 2(x - 4) = (x + 2) (x - 4)
To find zeroes, put p(x) = 0
\(\Rightarrow\) x - 4 = 0 or x + 2 = 0 \(\Rightarrow\) x = 4 or x = -2
Hence, zeroes of the given polynomial are -2 and 4.
Verification
Here, sum of the zeroes = -2 + 4 = 2 = -(-2/1)
\(=-\frac{Coefficient \quad of \quad x}{Coefficient \quad of \quad x^{2}}\)
and product of the zeroes = -2 \(\times\) 4 = -8 = -8/1
\(=\frac{Constant \quad term}{Coefficient \quad of \quad x^{2}}\)
So, the relationship between the zeroes and its coefficients is verified.

\(2,-\frac { 2 }{ 5 } \)

Let the polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+\beta=\sqrt{2}=\frac{-b}{a}\) 
\(\alpha \beta=\frac{1}{3}=\frac{c}{a}\) 
If a = 3, then \(b=\sqrt{2}\) and  c = 1/3
Therefore, the quadratic polynomial is \(3 x^{2}-3 \sqrt{2} x+1\).

Given, sum of zeroes = 1/4 and product of zeroes = -1
Then, the quadratic polynomial
=x² - (sum of zeroes)x + product of zeroes
\(=x^{2}-\left ( \frac{1}{4} \right )x-1=x^{2}-\frac{x}{4}-1=\frac{4x^{2}-x-4}{4}\)
We can consider 4x² - x - 4 as required quadratic polynomial because it will also staisfy the given conditions.

We have
x² + 7x + 10 = (x + 2)(x + 5)
So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or x = –5. Therefore, the zeroes of x² + 7x + 10 are – 2 and – 5. Now,
sum of zeroes = \(-2+(-5)=-(7)=\frac{-(7)}{1}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\) 
product of zeroes = \((-2) \times(-5)=10=\frac{10}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

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
\(=\left( \frac { 7 }{ 2 } \right) ^{ 2 }-3=\frac { 49 }{ 4 } -\frac { 3 }{ 1 } =\frac { 37 }{ 4 } \)

Given zeroes of polynomial are \(-\sqrt{\frac{5}{2}} \text { and } \sqrt{\frac{5}{2}}\)

∴ Sum of zeroes = \(-\sqrt{\frac{5}{2}}+\sqrt{\frac{5}{2}}=0\) and product of zeroes = \(-\sqrt{\frac{5}{2}} \cdot \sqrt{\frac{5}{2}}=-\frac{5}{2}\)

A quadratic polynomial with two zeroes is given by k(x² - (α + β)x + αβ).

\(\therefore k\left(x^2-(0) x+\left(\frac{-5}{2}\right)\right)=k\left(x^2-\frac{5}{2}\right)\)

For k = 8, the quadratic polynomial is \(8\left(x^2-\frac{5}{2}\right)=8 x^2-20\)

Given zeroes of polynomial are \(\frac{2}{5} \text { and } \frac{-1}{5}\)

Let \(\alpha=\frac{2}{5} \text { and } \beta=-\frac{1}{5}\)

∴ Sum of zeroes = \(\alpha+\beta=\frac{2}{5}-\frac{1}{5}=\frac{1}{5}\)

Product of zeroes = \(\alpha \beta=\frac{2}{5} \times\left(-\frac{1}{5}\right)=-\frac{2}{25}\)

∴ Required quadratic polynomial

= x² - (α + β)x + αβ

\(=x^2-\frac{1}{5} x+\left(-\frac{2}{25}\right)=\frac{25 x^2-5 x-2}{25}\)

We can consider 25x² - 5x - 2 as required quadratic polynomial because it will also satisfy the given condition.

Let p(x) = x² - 1

Put p(x) = 0 ⇒ x² - 1 = 0

⇒ x² = 1 ⇒ \(x= \pm 1\)

So, the zeroes are 1 and - 1.

Let α = 1 and β = -1

⇒ (α + β) = 1 + (-1) = 0

Let α = 2 and β = -3

Then, the quadratic polynomial will be

x² - (α + β)x + αβ = x² - (2 - 3)x + 2(-3)

= x² - (-1)x - 6 = x² + x - 6

On comparing this polynomial with x² + (a + 1)x + b, we get

a + 1 = 1 and b = -6

⇒ a = 0 and b = -6

In the given figure, the graph of a polynomial p(x) cuts X-axis at two distinct points. Hence, there are only two zeroes of the polynomial p(x).

Let f(x) = x² - 5x + 6

By splitting the middle term, we get

f(x) = x² - 3x - 2x + 6 = x(x - 3) - 2(x - 3)

⇒ f(x) = (x - 3)(x - 2)

On putting f(x) = 0, we get

x - 3 = 0 or x - 2 = 0

⇒ x = 3 or x = 2

Let p(x) = 16x² - 9

To find zeroes of p(x), put p(x) = 0

∴ 16x² - 9 = 0 ⇒ 16x² = 9

⇒ \(x^2=\frac{9}{16} \Rightarrow x= \pm \frac{3}{4}\)

So, the zeroes are \(\frac{3}{4} \text { and } \frac{-3}{4}\).

We know that for any quadratic polynomial ax² + bx + c, a ≠ 0, the graph of the corresponding equation y = ax + bx + c has one of the two shapes either open upwards like \(\cup\) or open downwards like \(\cap\) depending on whether a > 0 or a < 0. So, option (d) cannot be possible. Also, the curve of a quadratic polynomial crosses the X axis atmost at two points, but in option (d) the curve crosses the X-axis at three points, so it does not represent the quadratic polynomial.

Let f(x) = x² - 15
\(\Rightarrow f(x)=(x-\sqrt{15})(x+\sqrt{15}) \quad\left[\because a^2-b^2=(a-b)(a+b)\right]\)
On putting f(x) = 0, we get
\(\begin{aligned} & (x-\sqrt{15})(x+\sqrt{15})=0 \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow x-\sqrt{15}=0 \text { or } x+\sqrt{15}=0 \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad x=\sqrt{15} \text { or } x=-\sqrt{15} \end{aligned}\)
Thus, the zeros of given polynomial are \(\alpha=\sqrt{15}\) and \(\beta=-\sqrt{15}\)
Verification
Here, the sum of zeroes, \(\alpha+\beta=\sqrt{15}-\sqrt{15}=0\)
\(=-\frac{0}{1}=-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)
and product of zeroes,
\(\begin{aligned} \alpha \beta=\sqrt{15} \times(-\sqrt{15}) & =-15=\frac{-15}{1} \end{aligned}\)
\(\begin{aligned} & =\frac{\text { Constant term }}{\text { Coefficient of } x^2} \end{aligned}\)
So, the relationship between the zeroes and the coefficients is verified.

3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x - 4) +1(3x - 4)
=(3x - 4)(x +1)
The zeroes of the polynomial are {4/3, -1}
Relationship between the zeroes and the coefficient of the polynomial:
Also sum of the zeroes = \(\frac{4}{3}-1=\frac{4-3}{3}=\frac{1}{3}\) 
Also product of the zeroes = \(\frac{4}{3} x-1=-\frac{4}{3}\) 
Hence verified.

Let p(t) = t² - 15 = t² - \(\left ( \sqrt{15} \right )^{2}\)
= (t - \(\sqrt{15}\))(t + \(\sqrt{15}\))     [\(\because\) a² - b² = (a - b) (a + b)]
To find zeroes, put p(t) = 0
\(\Rightarrow (t-\sqrt{15})(t+\sqrt{15})=0\)
\(\Rightarrow t-\sqrt{15}=0\) or t + \(\sqrt{15}\) = 0 \(\Rightarrow\) t = \(\sqrt{15}\) or t = -\(\sqrt{15}\)
Hence, zeroes of the given polynomial are -\(\sqrt{15}\) and \(\sqrt{15}\).
Verification
Hence, sum of zeroes = -\(\sqrt{15}\) + \(\sqrt{15}\) = 0 = -(0/1)
\(=-\frac{Coefficient \quad of \quad t}{Coefficient \quad of \quad t^{2}}\)
and product of zeroes = -\(\sqrt{15}\) \(\times\)\(\sqrt{15}\)= -15 = \(\frac{-15}{1}\)
\(=\frac{Constant \quad term}{Coefficient \quad of \quad t^{2}}\)
So, the relationship between the zeroes and its coefficients is verified.

Let p(u) = 4u² + 8u = 4u(u+2)
To find zeroes, put p(u) = 0
\(\Rightarrow\) 4u(u+2) = 0 \(\Rightarrow\) u = 0 or u + 2 = 0 [\(\because\) 4 \(\neq\)0]
\(\Rightarrow\) u = 0 or u = -2
Hence, zeroes of the given polynonial are 0 and -2.
Verification
Here, sum of zeroes = 0 - 2 = -2 = -(8/4)
=-\(\frac{Coefficient \quad of \quad u}{Coefficient \quad of \quad u^{2}}\)
and product of zeroes
=0 \(\times\)-2 = 0 = (0/4) = \(\frac{Constant \quad term}{Coefficient \quad of \quad u^{2}}\)
so, the relationship between the zeroes and its coefficients is verified.

(i) (b): Graph of a quadratic polynomial is a parabolic in shape.
(ii) (c): Since the graph of the polynomial cuts the
x-axis at (-6,0) and (6, 0). So, the zeroes of polynomial are -6 and 6.
\(\therefore\) Required polynomial is p(x) = x² - (-6 + 6)x + (-6)(6) = x² - 36
(iii) (c) : We have, p(x) = x² - 36
Now, p( 6) = 62 - 36 = 36 - 36 = 0
(iv) (b): Letf (x) = x² + 2x - 3. Then,
\(\text { Sum of zeroes }=-\frac{\text { coefficient of } x}{\text { coefficient of } x^{2}}=-\frac{(2)}{1}=-2\)
(v) (d): The given polynomial is at²+ 5t + 3a Given, sum of zeroes = product of zeroes.
\(\Rightarrow \quad \frac{-5}{a}=\frac{3 a}{a} \Rightarrow a=\frac{-5}{3}\)