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

(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): The shape of the path of the soccer ball is a parabola.
(ii) (c): The axis of symmetry of the given curve is a line parallel to y-axis.
(iii) (a): The zeroes of the polynomial, represented in the given graph, are -2 and 7, since the curve cuts the x-axis at these points.
(iv) (d):A polynomial having zeroes -2 and -3 is \(p(x)=x^{2}-(-2-3) x+(-2)(-3)=x^{2}+5 x+6\)
(v) (c): We have \(f(x)=(x-3)^{2}+9\)
\(\text { Now, } 9=(x-3)^{2}+9 \)
\(\Rightarrow(x-3)^{2}=0 \Rightarrow x-3=0 \Rightarrow x=3\)

(i) (a): Since, the graph intersects the x-axis at two points, namely x = 8, -2. So, 8, - 2 are the zeroes of the given polynomial.
(ii) (b): The expression of the polynomial given in diagram is \(-x^{2}+6 x+16\)
(iii) (c) : Let \(p(x)=-x^{2}+6 x+16\)
\(\text { When } x=4, p(4)=-4^{2}+6 \times 4+16=24\)
(iv) (d): Let \(f(x)=-x^{2}+3 x-2\)
Now, consider \(f(x)=0 \Rightarrow-x^{2}+3 x-2=0\)
\(\begin{aligned} &\Rightarrow x^{2}-3 x+2=0 \Rightarrow(x-2)(x-1)=0\\ &\Rightarrow x=1,2 \text { are its zeroes. } \end{aligned}\)
(v) (b): Let a and \(\beta\) are the zeroes of the required polynomial.
Given \(\alpha + \beta = - 3\)
If \(\alpha\) = 4, then  \(\beta\)= -7
\(\therefore \quad \text { Representation of tunnel is }-x^{2}-3 x+28 \text { . }\)

(i) (c): For finding \(\alpha,\beta,\) \(\gamma\) consider p(x) = 0
\(\Rightarrow \quad x^{3}-18 x^{2}+95 x-150=0 \)
\(\Rightarrow \quad(x-3)\left(x^{2}-15 x+50\right)=0 \)
\(\Rightarrow \quad(x-3)(x-5)(x-10)=0 \Rightarrow x=10 \text { or } x=5 \text { or } x=3 \)
\(\text { Thus } \alpha=10, \beta=5 \text { and } \gamma=3\)
(ii) (d): Here \(\alpha=10, \beta=5 \text { and } \gamma=3\)
\(\therefore\) Sum of product of zeroes taken two at a time
\(\begin{array}{l} =\alpha \beta+\beta \gamma+\gamma \alpha=(10)(5)+(5)(3)+(3)(10) \\ =50+15+30=95 \end{array}\)
(iii) (a): Product of zeroes of polynomial p(x) = \(\alpha\beta\gamma\)
= (10) (5) (3) = 150
(iv) (b): We have \(p(x)=x^{3}-18 x^{2}+95 x-150\)
\(\begin{array}{l} \text { Now, } p(4)=4^{3}-18(4)^{2}+95(4)-150 \\ =64-288+380-150=6 \end{array}\)
(v) (d): \(g(x)=x^{3}-(\alpha+\beta+\gamma) x^{2} +(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma \)
\(\Rightarrow g(x)=x^{3}-3 x^{2}-16 x-(-48)=x^{3}-3 x^{2}-16 x+48\)