Class 10: Polynomials – CBSE Board Problems

$\displaystyle \textbf{Question 1. }\text{Which of the following statements is true for a polynomial } \\ p(x)\text{ of degree }3? \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{(a) }p(x)\text{ has at most two distinct zeroes.}$
$\displaystyle \text{(b) }p(x)\text{ has at least two distinct zeroes.}$
$\displaystyle \text{(c) }p(x)\text{ has exactly three distinct zeroes.}$
$\displaystyle \text{(d) }p(x)\text{ has at most three distinct zeroes.}$
$\displaystyle \text{Answer:}$
$\displaystyle (d)\ \therefore \text{A polynomial of degree }n\text{ has at most }n\text{ distinct roots.}$
$\\$

$\displaystyle \textbf{Question 2. }\alpha\text{ and }\beta\text{ are zeroes of a quadratic polynomial}$
$\displaystyle px^{2}+qx+1.\text{ Form a quadratic polynomial whose zeroes are }\frac{2}{\alpha}\text{ and }\frac{2}{\beta}. \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given quadratic polynomial is }px^{2}+qx+1\text{ and }\alpha,\beta\text{ are zeroes of this polynomial.}$
$\displaystyle \Rightarrow \alpha+\beta=-\frac{q}{p},\ \alpha\beta=\frac{1}{p}$
$\displaystyle \text{Now, we have to form a quadratic polynomial whose zeroes are }\frac{2}{\alpha}\text{ and }\frac{2}{\beta}.$
$\displaystyle \text{Sum of zeroes}=\frac{2}{\alpha}+\frac{2}{\beta}=\frac{2(\alpha+\beta)}{\alpha\beta}$
$\displaystyle =\frac{2\left(-\frac{q}{p}\right)}{\frac{1}{p}}=-2q$
$\displaystyle \text{Product of zeroes}=\frac{2}{\alpha}\cdot\frac{2}{\beta}=\frac{4}{\alpha\beta}=\frac{4}{\frac{1}{p}}=4p$
$\displaystyle \text{Now, formation of polynomial is given by}$
$\displaystyle x^{2}-(\text{Sum of zeroes})x+\text{Product of zeroes}$
$\displaystyle =x^{2}-(-2q)x+4p$
$\displaystyle =x^{2}+2qx+4p$
$\\$

$\displaystyle \textbf{Question 3. }\text{The zeroes of a polynomial }x^{2}+px+q\text{ are twice the zeroes} \\ \text{of the polynomial }4x^{2}-5x-6.\text{ The value of }p\text{ is:} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) }-\frac{5}{2} \qquad \text{(b) }\frac{5}{2} \qquad \text{(c) }-5 \qquad \text{(d) }10$
$\displaystyle \text{Answer:}$
$\displaystyle (a)\ \text{Let }f(x)=4x^{2}-5x-6$
$\displaystyle \text{Now, for zeroes of }f(x)$
$\displaystyle f(x)=0$
$\displaystyle \Rightarrow 4x^{2}-5x-6=0$
$\displaystyle \Rightarrow (4x+3)(x-2)=0$
$\displaystyle \Rightarrow x=-\frac{3}{4}\ \text{or}\ x=2$
$\displaystyle \text{Now, zeroes of polynomial }x^{2}+px+q\text{ are:}$
$\displaystyle 2\times\left(-\frac{3}{4}\right)\text{ and }2\times2\text{ i.e. }-\frac{3}{2}\text{ and }4$
$\displaystyle \text{Now, sum of zeroes of polynomial }x^{2}+px+q=-p$
$\displaystyle \Rightarrow -\frac{3}{2}+4=-p$
$\displaystyle \Rightarrow p=-\frac{5}{2}$
$\\$

$\displaystyle \textbf{Question 4. }\text{If the sum of zeroes of the polynomial } \\ p(x)=2x^{2}-k\sqrt{2}x+1\text{ is }\sqrt{2},\text{ then value of }k\text{ is:} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) }\sqrt{2} \qquad \text{(b) }2 \qquad \text{(c) }2\sqrt{2} \qquad \text{(d) }\frac{1}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle (d)\ \text{As, }p(x)=2x^{2}-k\sqrt{2}\,x+1$
$\displaystyle \text{Sum of zeroes}=\sqrt{2}$
$\displaystyle \Rightarrow \frac{-\text{coefficient of }x}{\text{coefficient of }x^{2}}=\sqrt{2}$
$\displaystyle \Rightarrow \frac{-(-k\sqrt{2})}{2}=\sqrt{2}$
$\displaystyle \Rightarrow k=2$
$\\$

$\displaystyle \textbf{Question 5. }\text{Directions: In the following question, a statement of assertion }(A)\text{ is followed}$
$\displaystyle \text{by a statement of reason }(R).\text{ Choose the correct answer out of the following choices.}$
$\displaystyle \text{Assertion }(A):\text{ If the graph of a polynomial touches }x\text{-axis at only one point, then} \\ \text{the polynomial cannot be a quadratic polynomial.}$
$\displaystyle \text{Reason }(R):\text{ A polynomial of degree }n(n>1)\text{ can have at most }n\text{ zeroes.} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) Both Assertion }(A)\text{ and Reason }(R)\text{ are true and Reason }(R)\text{ is the correct} \\ \text{explanation of Assertion }(A).$
$\displaystyle \text{(b) Both Assertion }(A)\text{ and Reason }(R)\text{ are true and Reason }(R)\text{ is not the} \\ \text{correct explanation of Assertion }(A).$
$\displaystyle \text{(c) Assertion }(A)\text{ is true but Reason }(R)\text{ is false.}$
$\displaystyle \text{(d) Assertion }(A)\text{ is false but Reason }(R)\text{ is true.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A quadratic polynomial can touch the }x\text{-axis at exactly one point.}$
$\displaystyle \text{For example, }y=(x-1)^2\text{ touches the }x\text{-axis at }x=1.$
$\displaystyle \therefore \text{Assertion (A) is false.}$
$\displaystyle \text{A polynomial of degree }n(n>1)\text{ can have at most }n\text{ zeroes.}$
$\displaystyle \therefore \text{Reason (R) is true.}$
$\displaystyle \therefore \text{Assertion (A) is false but Reason (R) is true.}$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 6. }\text{The zeroes of the polynomial }3x^{2}+11x-4\text{ are:} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{(a) }\frac{1}{2},-4 \qquad \text{(b) }\frac{1}{4},-3$
$\displaystyle \text{(c) }\frac{1}{3},-4 \qquad \text{(d) }\frac{1}{3},4$
$\displaystyle \text{Answer:}$
$\displaystyle (c)\ \text{Let }p(x)=3x^{2}+11x-4$
$\displaystyle \text{Now, zeroes of }p(x)\text{ are given by solving }p(x)=0$
$\displaystyle \Rightarrow 3x^{2}+11x-4=0$
$\displaystyle \Rightarrow (3x-1)(x+4)=0$
$\displaystyle \Rightarrow x=\frac{1}{3},\ -4$
$\\$

$\displaystyle \textbf{Question 7. }\text{If }\alpha,\beta\text{ are zeroes of the polynomial }x^{2}-1,\text{ then value of } \\ (\alpha+\beta)\text{ is:} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }2 \qquad \text{(b) }1 \qquad \text{(c) }-1 \qquad \text{(d) }0$
$\displaystyle \text{Answer:}$
$\displaystyle (d)\ p(x)=x^{2}-1$
$\displaystyle \Rightarrow a=1,\ b=0,\ c=-1$
$\displaystyle \therefore \alpha+\beta=\frac{-b}{a}=\frac{0}{1}=0$
$\\$

$\displaystyle \textbf{Question 8. }\text{If }\alpha,\beta\text{ are the zeroes of the polynomial } \\ p(x)=4x^{2}-3x-7,\text{ then }\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\text{ is equal to:} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\frac{7}{3} \qquad \text{(b) }-\frac{7}{3}$
$\displaystyle \text{(c) }\frac{3}{7} \qquad \text{(d) }-\frac{3}{7}$
$\displaystyle \text{Answer:}$
$\displaystyle (d)\ p(x)=4x^{2}-3x-7$
$\displaystyle \text{Here, }a=4,\ b=-3,\ c=-7$
$\displaystyle \therefore \alpha,\beta\text{ are zeroes of }p(x)$
$\displaystyle \therefore \alpha+\beta=\frac{-b}{a}=\frac{3}{4},\ \alpha\beta=\frac{c}{a}=-\frac{7}{4}$
$\displaystyle \text{Now, }\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha\beta}$
$\displaystyle =\frac{\frac{3}{4}}{-\frac{7}{4}}=-\frac{3}{7}$
$\\$

$\displaystyle \textbf{Question 9. }\text{If one zero of the polynomial }p(x)=6x^{2}+37x-(k-2) \\ \text{is reciprocal of the other, then find the value of }k. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \ p(x)=6x^{2}+37x-(k-2)$
$\displaystyle \text{Here, }a=6,\ b=37,\ c=-(k-2)=2-k$
$\displaystyle \text{Let one zero }=\alpha$
$\displaystyle \therefore \text{Other zero}=\frac{1}{\alpha}$
$\displaystyle \text{Product of zeroes}=\frac{c}{a}$
$\displaystyle \Rightarrow \alpha\times\frac{1}{\alpha}=\frac{2-k}{6}$
$\displaystyle \Rightarrow 1=\frac{2-k}{6}$
$\displaystyle \Rightarrow k=-4$
$\\$

$\displaystyle \textbf{Question 10. }\text{While playing in a garden, Samaira saw a honey-comb and} \\ \text{asked her mother what is that. Her mother } \text{replied that it's a honeycomb made} \\ \text{by honey bees to store honey. Also, she told her that the shape of the honeycomb } \\ \text{formed is a } \text{mathematical structure. The mathematical representation of the} \\ \text{honeycomb is shown in the graph.} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{Based on the above information, answer the following questions:}$
$\displaystyle \text{(i) How many zeroes are there for the polynomial represented by the graph given?}$
$\displaystyle \text{(ii) Write the zeroes of the polynomial.}$
$\displaystyle \text{(iii) If the zeroes of the polynomial }x^{2}+(a+1)x+b\text{ are }2\text{ and }-3,\text{ then determine the} \\ \text{values of }a\text{ and }b.$
$\displaystyle \text{OR}$
$\displaystyle \text{If the square of difference of the zeroes of the polynomial }x^{2}+px+45\text{ is }144, \\ \text{ then find the value of }p.$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \text{The shape of the given graph is a parabola that intersects the }x\text{-axis at }2\text{ points.}$
$\displaystyle \therefore \text{The given polynomial has }2\text{ zeroes.}$
$\displaystyle \text{(ii) The zeroes of the given polynomial are }-7\text{ and }7.$
$\displaystyle \text{(iii) Let }p(x)=x^{2}+(a+1)x+b$
$\displaystyle \text{Now, }2\text{ and }-3\text{ are the zeroes of }p(x),\ \text{then }p(2)=0$
$\displaystyle \Rightarrow (2)^{2}+(a+1)(2)+b=0$
$\displaystyle \Rightarrow 2a+b=-6 \qquad (1)$
$\displaystyle \text{Also, }p(-3)=0$
$\displaystyle \Rightarrow (-3)^{2}+(a+1)(-3)+b=0$
$\displaystyle \Rightarrow -3a+b=-6 \qquad (2)$
$\displaystyle \text{Solving }(1)\text{ and }(2),\ \text{we get}$
$\displaystyle a=0\text{ and }b=-6$
$\displaystyle \text{OR}$
$\displaystyle \text{Let }p(x)=x^{2}+px+45$
$\displaystyle \text{Suppose }\alpha\text{ and }\beta\text{ are zeroes of }p(x).$
$\displaystyle \text{Now, }(\alpha-\beta)^{2}=144$
$\displaystyle \Rightarrow (\alpha+\beta)^{2}-4\alpha\beta=144$
$\displaystyle \Rightarrow (-p)^{2}-4\times45=144$
$\displaystyle \Rightarrow p^{2}=144+180$
$\displaystyle \Rightarrow p^{2}=324$
$\displaystyle \Rightarrow p=\pm18$

$\displaystyle \textbf{Question 11. }\text{In figure, the graph of a polynomial }P(x)\text{ is shown. The number} \\ \text{of zeroes of }P(x)\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }1 \qquad \text{(b) }2$
$\displaystyle \text{(c) }3 \qquad \text{(d) }4$
$\displaystyle \text{Answer:}$
$\displaystyle (c)\ \text{Since, the curve cuts the }x\text{-axis at }3\text{ places.}$
$\displaystyle \therefore \text{Number of zeroes}=3$
$\\$

$\displaystyle \textbf{Question 12. }\text{The graph of a polynomial }P(x)\text{ cuts the }x\text{-axis at }3 \\ \text{ points and touches it at }2\text{ other points. The number of zeroes of }P(x)\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }1 \qquad \text{(b) }2$
$\displaystyle \text{(c) }3 \qquad \text{(d) }5$
$\displaystyle \text{Answer:}$
$\displaystyle (d)\ \text{When the graph cuts or touches the }x\text{-axis, }p(x)=0.$
$\displaystyle \text{Since, the graph cuts or touches at }5\text{ places.}$
$\displaystyle \therefore \text{The number of zeroes}=5$
$\\$

$\displaystyle \textbf{Question 13. }\text{If }x-1\text{ is a factor of the polynomial }p(x)=x^{3}+ax^{2}+2b\text{ and } \\ a+b=4,\text{ then} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }a=5,b=-1 \qquad \text{(b) }a=9,b=-5$
$\displaystyle \text{(c) }a=7,b=-3 \qquad \text{(d) }a=3,b=1$
$\displaystyle \text{Answer:}$
$\displaystyle (b)\ \text{Since }x-1\text{ is a factor of the polynomial }p(x)=x^{3}+ax^{2}+2b$
$\displaystyle \therefore (1)^{3}+a(1)^{2}+2b=0$
$\displaystyle \Rightarrow a+2b=-1 \qquad (i)$
$\displaystyle \text{and }a+b=4 \qquad (ii)$
$\displaystyle \text{Subtracting }(i)\text{ from }(ii),\ b=5$
$\displaystyle \text{Now, putting }b=5\text{ in }(ii),$
$\displaystyle a=9$
$\\$

$\displaystyle \textbf{Question 14. }\text{A quadratic polynomial, the product and sum of whose zeroes are } \\ 5\text{ and }8\text{ respectively is} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }k[x^{2}-8x+5] \qquad \text{(b) }k[x^{2}+8x+5]$
$\displaystyle \text{(c) }k[x^{2}-5x+8] \qquad \text{(d) }k[x^{2}+5x+8]$
$\displaystyle \text{Answer:}$
$\displaystyle (a)\ \text{We know that if }\alpha\text{ and }\beta\text{ are zeroes of a quadratic polynomial,}$
$\displaystyle \text{then }p(x)=k\left[x^{2}-(\alpha+\beta)x+\alpha\beta\right]$
$\displaystyle \text{Here, }\alpha\beta=5\text{ and }\alpha+\beta=8$
$\displaystyle \therefore \text{Quadratic polynomial}=k(x^{2}-8x+5)$
$\\$

$\displaystyle \textbf{Question 15. }\text{If }\alpha,\beta\text{ are the zeros of the quadratic polynomial } \\ p(x)=x^{2}-(k+6)x+2(2k-1),\text{ then the value of }k,$
$\displaystyle \text{if }\alpha+\beta=\frac{1}{2}\alpha\beta,\text{ is} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }-7 \qquad \text{(b) }7$
$\displaystyle \text{(c) }-3 \qquad \text{(d) }3$
$\displaystyle \text{Answer:}$
$\displaystyle (b)\ \text{According to the question,}$
$\displaystyle k+6=\frac{1}{2}\times2(2k-1)$
$\displaystyle \Rightarrow k+6=2k-1$
$\displaystyle \Rightarrow k=7$
$\\$

$\displaystyle \textbf{Question 6. }\text{If one of the zeroes of the quadratic polynomial }x^{2}+3x+k\text{ is }2, \\ \text{ then the value of }k\text{ is} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{(a) }10 \qquad \text{(b) }-10$
$\displaystyle \text{(c) }-7 \qquad \text{(d) }-2$
$\displaystyle \text{Answer:}$
$\displaystyle (b)\ \text{As one zero is }2,$
$\displaystyle \Rightarrow (2)^{2}+3\times2+k=0$
$\displaystyle \Rightarrow k=-4-6=-10$
$\displaystyle \text{Alternatively, let other zero be }\alpha.$
$\displaystyle \therefore \alpha+2=-3\text{ and }2\alpha=k$
$\displaystyle \Rightarrow \alpha=-5\text{ and }2\times(-5)=k$
$\displaystyle \Rightarrow k=-10$
$\\$

$\displaystyle \textbf{Question 17. }\text{The quadratic polynomial, the sum of whose zeroes is }-5\text{ and their} \\ \text{product is }6,\text{ is} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{(a) }x^{2}+5x+6 \qquad \text{(b) }x^{2}-5x+6$
$\displaystyle \text{(c) }x^{2}-5x-6 \qquad \text{(d) }-x^{2}+5x+6$
$\displaystyle \text{Answer:}$
$\displaystyle (a)\ \text{Sum of zeroes}=-5,\ \text{product}=6$
$\displaystyle \text{Polynomial is, }x^{2}-(\text{sum of zeroes})x+\text{product of zeroes}$
$\displaystyle \Rightarrow x^{2}-(-5)x+6=x^{2}+5x+6$
$\\$

$\displaystyle \textbf{Question 18. }\text{Find a quadratic polynomial whose zeroes are reciprocals of the zeroes}$
$\displaystyle \text{of the polynomial } f(x)=ax^{2}+bx+c,\ a\neq 0,\ c\neq 0. \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given polynomial }f(x)=ax^{2}+bx+c,\ a\neq0,\ c\neq0$
$\displaystyle \text{Let zeroes be }\alpha,\beta$
$\displaystyle \alpha+\beta=-\frac{b}{a},\ \alpha\beta=\frac{c}{a} \qquad (i)$
$\displaystyle \text{For required polynomial, reciprocal zeroes are }\frac{1}{\alpha},\frac{1}{\beta}$
$\displaystyle \text{Sum}=\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}$
$\displaystyle =\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac{b}{c}=-\frac{B}{A}$
$\displaystyle \text{Product}=\frac{1}{\alpha}\cdot\frac{1}{\beta}=\frac{1}{\alpha\beta}=\frac{a}{c}=\frac{C}{A}$
$\displaystyle \text{Polynomial is }Ax^{2}+Bx+C$
$\displaystyle \text{Here, }A=c,\ B=b,\ C=a$
$\displaystyle \therefore \text{Required polynomial is }cx^{2}+bx+a$
$\\$

$\displaystyle \textbf{Question 19. }\text{Find the zeroes of the quadratic polynomial }7y^{2}-\frac{11}{3}y-\frac{2}{3}\text{ and verify the }$
$\displaystyle \text{relationship between the zeroes and the coefficients.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here }p(y)=7y^{2}-\frac{11}{3}y-\frac{2}{3}$
$\displaystyle \text{For zeroes of }p(y),\ p(y)=0$
$\displaystyle \Rightarrow 7y^{2}-\frac{11}{3}y-\frac{2}{3}=0$
$\displaystyle \Rightarrow 21y^{2}-11y-2=0$
$\displaystyle \Rightarrow (7y+1)(3y-2)=0$
$\displaystyle \Rightarrow (7y+1)(y-\frac{2}{3})=0$
$\displaystyle \Rightarrow y=-\frac{1}{7},\ \frac{2}{3}$
$\displaystyle \therefore \text{Zeroes are }-\frac{1}{7}\text{ and }\frac{2}{3}$
$\displaystyle \text{Also }a=7,\ b=-\frac{11}{3},\ c=-\frac{2}{3}$
$\displaystyle \text{Sum of zeroes}=-\frac{1}{7}+\frac{2}{3}=\frac{-3+14}{21}=\frac{11}{21}$
$\displaystyle \text{Also }-\frac{b}{a}=-\frac{-11/3}{7}=\frac{11}{21}$
$\displaystyle \therefore \text{Sum of zeroes}=-\frac{b}{a}$
$\displaystyle \text{Product of zeroes}=-\frac{1}{7}\times\frac{2}{3}=-\frac{2}{21}$
$\displaystyle \text{Also }\frac{c}{a}=\frac{-2/3}{7}=-\frac{2}{21}$
$\displaystyle \therefore \text{Product of zeroes}=\frac{c}{a}$
$\\$

$\displaystyle \textbf{Question 20. }\text{Find the quadratic polynomial, sum and product of whose zeroes are }$
$\displaystyle -1\text{ and }-20\text{ respectively.} \text{Also, find the zeroes of the polynomial so obtained.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Sum of zeroes}=-1,\ \text{Product of zeroes}=-20$
$\displaystyle \text{Given, quadratic polynomial}=x^{2}-(\text{sum of zeroes})x+\text{product of zeroes}$
$\displaystyle =x^{2}-(-1)x+(-20)$
$\displaystyle =x^{2}+x-20$
$\displaystyle \text{On factorising by splitting the middle term,}$
$\displaystyle x^{2}+x-20=0$
$\displaystyle \Rightarrow x^{2}+5x-4x-20=0$
$\displaystyle \Rightarrow x(x+5)-4(x+5)=0$
$\displaystyle \Rightarrow (x-4)(x+5)=0$
$\displaystyle \therefore \text{Other zeroes are }4\text{ and }-5$
$\\$

$\displaystyle \textbf{Question 21. }\text{Find the zeroes of the quadratic polynomial }3x^{2}-2\text{ and verify the }$
$\displaystyle \text{relationship between the zeroes and the coefficients.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic polynomial is }3x^{2}-2$
$\displaystyle \text{Consider }p(x)=3x^{2}-2$
$\displaystyle \text{For zeroes of polynomial }p(x),\ p(x)=0$
$\displaystyle \Rightarrow 3x^{2}-2=0$
$\displaystyle \Rightarrow 3x^{2}=2$
$\displaystyle \Rightarrow x^{2}=\frac{2}{3}$
$\displaystyle \Rightarrow x=\pm\sqrt{\frac{2}{3}}$
$\displaystyle \therefore \text{Zeroes of polynomial are }\sqrt{\frac{2}{3}}\text{ and }-\sqrt{\frac{2}{3}}$
$\displaystyle \text{Here }\alpha=\sqrt{\frac{2}{3}}\text{ and }\beta=-\sqrt{\frac{2}{3}}$
$\displaystyle \text{Hence, sum of zeroes}=\alpha+\beta=0$
$\displaystyle \text{Product of zeroes}=\alpha\beta=-\frac{2}{3}$
$\displaystyle \text{Also, from the polynomial }p(x)=3x^{2}-2,$
$\displaystyle \text{Sum of zeroes}=-\frac{b}{a}=-\frac{0}{3}=0$
$\displaystyle \text{Product of zeroes}=\frac{c}{a}=-\frac{2}{3}$
$\displaystyle \text{This verifies the relation.}$
$\\$

$\displaystyle \textbf{Question 22. }\text{Quadratic polynomial }2x^{2}-3x+1\text{ has zeroes as }\alpha\text{ and }\beta.\text{ Now }$
$\displaystyle \text{form a quadratic polynomial whose zeroes are }3\alpha\text{ and }3\beta. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \alpha\text{ and }\beta\text{ are zeroes of the polynomial }2x^{2}-3x+1$
$\displaystyle \Rightarrow \alpha+\beta=\frac{-(-3)}{2}=\frac{3}{2}$
$\displaystyle \alpha\beta=\frac{1}{2}$
$\displaystyle \text{Now, zeroes of the required polynomial are }3\alpha\text{ and }3\beta$
$\displaystyle S=3\alpha+3\beta=3(\alpha+\beta)=3\left(\frac{3}{2}\right)=\frac{9}{2}$
$\displaystyle P=(3\alpha)(3\beta)=9\alpha\beta=9\times\frac{1}{2}=\frac{9}{2}$
$\displaystyle \text{Now, required polynomial is }k(x^{2}-Sx+P)$
$\displaystyle =k\left(x^{2}-\frac{9}{2}x+\frac{9}{2}\right)$
$\displaystyle =\frac{k}{2}(2x^{2}-9x+9),\ \text{where }k\text{ is any constant.}$
$\\$