(i) (c) :The shape represents a quadratic polynomial.
(ii) (c): Since, the graph of polynomial cuts the x-axis at (-2, 0) and (4, 0). So, the polynomial has 2 zeroes.
(iii) (a) : The zeroes of the polynomial are -2 and 4.
(iv) (b): Required polynomial is \(p(x)=x^{2}-(-2+4) x+(-2)(4)=x^{2}-2 x-8\)
(v) (c): Consider, p(x) = -5
\(\Rightarrow \quad x^{2}-2 x-8=-5 \Rightarrow x^{2}-2 x-3=0\)
\(\Rightarrow(x-3)(x+1)=0 \Rightarrow x=-1 \text { or } x=3\)
\(\text { So, at } x=3 \text { and at } x=-1, p(x)=-5 \text { . }\)

(i) (b): Since, the graph intersects the x-axis at two points, namely x = -4, 7
So, -4, 7 are the zeroes of the polynomial.
(ii) (d): p(x) = -x² + 3x + 28
(iii) (a) : \(\text { Product of zeroes }=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
\(\therefore \quad \text { Required product of zeroes }=\frac{28}{-1}=-28\)
(iv) (c): We have \(9 x^{2}-5 =(3 x)^{2}-(\sqrt{5})^{2} =(3 x-\sqrt{5})(3 x+\sqrt{5}) \)
\(\therefore x=\frac{\sqrt{5}}{3} \text { or } \frac{-\sqrt{5}}{3}\)
(v) (b): Here, \(f(x)=x^{2}-13 x+1\)
\(\therefore \quad f(4)=4^{2}-13(4)+1=16-52+1=-35\)

(i) (b): Given, a and \(\beta\) are the zeroes of 
\(p(x)=x^{2}-24 x+128\)
\(\text { Putting } p(x)=0 \text { , we get }\)
\( x^{2}-8 x-16 x+128=0 \)
\(\Rightarrow x(x-8)-16(x-8)=0 \)
\(\Rightarrow (x-8)(x-16)=0 \Rightarrow x=8 \text { or } x=16 \)
\(\therefore \alpha=8, \beta=16\)
(ii) (c) : \(\alpha+\beta+\alpha \beta =8+16+(8)(16) =24+128=152 \)
(iii) (d) : \(p(2)=2^{2}-2 4(2)+128=4-48+128=84\)
(iv) (a): Since a and \(\beta\) are zeroes of \(x^{2}+x-2\)
\(\therefore \quad \alpha+\beta=-1 \text { and } \alpha \beta=-2 \)
\(\text { Now, } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-1}{-2}=\frac{1}{2}\)
(v) (c): Sum of zeroes \(=\frac{-2}{k}\)
Product of zeroes \(=\frac{3 k}{k}=3\)
According to question, we have \(\frac{-2}{k}=3\)
\(\Rightarrow \quad k=\frac{-2}{3}\)

(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): 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.
\(\Rightarrow \quad \frac{-5}{a}=\frac{3 a}{a} \Rightarrow a=\frac{-5}{3}\)