(i) Assume the values of zeroes of q(x) as \((-\alpha), \alpha\) and \(\beta\).
Write the sum of zeroes as
\(-\alpha+\alpha+\beta=2 \Rightarrow \beta=2\)
Writes the equation for the sum of the products of zeroes taken two at a time as:
\((-\alpha \times \alpha)+\alpha \beta+(-\alpha) \beta=-9\)
\( \Rightarrow \quad-\alpha^2-\alpha \beta+\beta \alpha=-9 \Rightarrow \alpha^2=9 \Rightarrow \alpha= \pm 3\)
Thus, 3 zeroes of q(x) are (-3), 3 and 2 .
(ii) Product of zeroes of given polynomial is \( (-\alpha)(\alpha) \beta  =-k \)
\(\Rightarrow  (-3)(3)(2)  =-k \Rightarrow-18=-k \)
\(\therefore  k  =18\)

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.

We have, \(\alpha+\beta=3\) and \(\alpha \beta=-2\) Now, required polynomial is given by
\(x^{2}-\left(\frac{2 \alpha}{\beta}+\frac{2 \beta}{\alpha}\right) x+\left(\frac{2 \alpha}{\beta}\right) \cdot\left(\frac{2 \beta}{\alpha}\right)\) 
\(=x^{2}-2\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right) x+4\) 
\(=x^{2} - 2\left(\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}\right) x+4\) 
Ans.x²+ 13x+ 4

Given, 2 and -3 are the zeros of x²+(a+1)x+b.
On putting x=2, we get 
(2)²+(a+1)2+b=0
⇒ 4+2a+2+b=0
⇒ 2a+b=6         ..(i)
Again putting x=-3, we get
(-3)²+(a+1)(-3)+b=0
⇒ 9-3a-3+b=0
⇒ -3a+b=-6  ..(ii)
On subtracting Eq. (ii) from Eq. (i), we get
5a=0⇒a=0
On putting a=0 in Eq. (ii), we get
-3(0)+b=-6⇒0+b=-6⇒b=6
Now, a+b=0-6=-6
Hence, the value of (a+b) is -6.

p(x)=x³-2x²+qx-r
Here,a =1,b = -2,c = q,d= -r
Sum of zeroes=\(\frac{-b}{a}\) ⇒ ∝+β+Y=\(\frac{-(-2)}{1}\)=2
⇒ 0+Y=2
⇒ Y=2...(i)
Also ∝β+βY+∝Y=\(\frac{c}{a}\) ⇒ ∝β+Υ(∝+β)=\(\frac{q}{r}\)
⇒ ∝β+Υx0=q ⇒ ∝β=q...(ii)
and ∝.β.Υ=\(\frac{-d}{a}\)
⇒ q.2=-(-r)
⇒ 2q=r

(i) Here, the graph of y = p(x) does not intersect the X-axis, so p(x) has no zero.
(ii) The number of zeros is one, as the graph of y = p(x) intersects the X-axis at one point only.
(iii) The number of zeros is three, as the graph of y = p(x) intersects the X-axis at three points.
(iv) The number of zeros is two, as the graph of y = p(x) intersects the X-axis at two points.
(v) The number of zeros is four, as the graph of y = p(x) intersects the X-axis at four points.
(vi) The number of zeros is three, as the graph of y = p(x) intersects the X-axis at three points.

Recall the identity a² - b² = (a - b)(a + b). Using it, we can write:
x² - 3 = (x - \(\sqrt{3}\))(x + \(\sqrt{3}\))
So, the value of x² - 3 is zero when x = \(\sqrt{3}\) or x = -\(\sqrt{3}\)
Therefore, the zeroes of x² - 3 are \(\sqrt{3}\) and -\(\sqrt{3}\)
Now,
sum of zeroes = \(\sqrt{3}\) - \(\sqrt{3}\) = 0 = \(\frac{-(Coefficient \quad of \quad x)}{Coefficient \quad of \quad x^{2}}\)
product of zeroes = (\(\sqrt{3}\))(-\(\sqrt{3}\)) = -3 = \(\frac{-3}{1}=\frac{Constant \quad term}{Coefficient \quad of \quad x^{2}}\)

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.

\(-\frac{1}{4}, \frac{1}{4}\) 
Let the quadratic polynomial be ax² + bx + c, and its zeroes be α + ß
Given  \(\alpha+\beta=-\frac{1}{4}=-\frac{b}{a}\) 
\(\alpha \beta=\frac{1}{4}=\frac{c}{a}\) 
If a = 4, b = 1, c = 1
Therefore, the quadratic polynomial is 4x² + x +1.

Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+\beta=-\frac{b}{a}=1\) 
\(\alpha \beta=\frac{c}{a}=1\) 
If a = 1, b = -1, and c = 1
Therefore, the quadratic polynomial is x² – 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.

Remainder=21x-3≠0
∴ g(x) is not a factor of p(x).

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

(i) (c) :Since the graph intersect the x-axis at 3 points, therefore the polynomial has 3 zeroes.
(ii) (d): Clearly the graph intersect the x-axis at x = -4, x = -2 and x = 1, therefore the zeroes are -4, -2 and 1. Now, the sum of zeroes = -4 - 2 + 1 = -5
(iii) (b): From the graph, it can be seen that When x = 0, then y = -8.
(iv) (b): Since there are 3 zeroes, therefore the graph represents a cubic polynomial.
(v) (a): The sum of product of zeroes taken two at a time = (-4)(-2) + (-2)(1) + (1)(-4) = 8 - 2 - 4 = 2

(i) (b): Since the graph of the polynomial intersect the x-axis at \(x=\frac{1}{2}, \frac{-7}{2}\), therefore required zeroes of the polynomial are \(\frac{1}{2} \text { and } \frac{-7}{2}\)
(ii) (d): \(\because \frac{1}{2} \text { and } \frac{-7}{2}\) are the zeroes of the polynomial.
So, at \(x=\frac{1}{2}, \frac{-7}{2}\) the value of the polynomial will be 0.
From options, required polynomial is
p(x) = -4x² - 12x + 7
(iii) (b) : we have, \(p(x)=-4 x^{2}-12 x+7\)
\(\therefore \quad p(3)=-4(3)^{2}-12(3)+7=-36-36+7=-65 \)
(iv) (c): Here \(f(x)=x^{2}+2 x-8 \text { and } \alpha, \beta \text { are its zeroes. }\)
\(\therefore \quad \alpha+\beta=-2 \text { and } \alpha \beta=-8 \)
\(\text { Now, } \alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2 \alpha^{2} \beta^{2} \)
\(=\left((\alpha+\beta)^{2}-2 \alpha \beta\right)^{2}-2(\alpha \beta)^{2} \)
\(=\left[(-2)^{2}-2(-8)\right]^{2}-2(-8)^{2} \)
\(=[4+16]^{2}-2(-8)^{2}=(20)^{2}-2(64) \)
\(=400-128=272\)
(v) (a): We have sum of zeroes = 0 and product of zeroes = \(\sqrt(7)\)
So, required polynomial .\(=k\left(x^{2}-0 \cdot x+\sqrt{7}\right) \)
\(=k\left(x^{2}+\sqrt{7}\right)\)

(i) (b): Since, the given graph is parabolic is shape, therefore it will represent a quadratic polynomial.
[\(\therefore\) Graph of quadratic polynomial is parabolic in shape 1
(ii) (c): Since, the graph cuts the x-axis at -1, 5. So the polynomial has 2 zeroes i.e., -1 and 5.
(iii) (a) : Sum of zeroes = -1 + 5 = 4
(iv) (c): Since a and β are zeroes of the given polynomial and β > a
\(\therefore\)a = - 1 and β = 5.
\(\therefore|8 \alpha+\beta|=|8(-1)+5|=|-8+5|=|-3|=3 .\)
(v) (d): Since the zeroes of the given polynomial are - 1 and 5.
\(\therefore\) Required polynomial p(x)
= k{ x² -(-1 + 5)x + (-1)(5)} = k(.x² - 4x - 5)
For k = -1, we get
p(x) = -.x² + 4x + 5, which is the required polynomial.