Question 1

The graphs of $y=p(x)$ are given in following figure, for some polynomials $p(x)$. Find the number of zeroes of $p(x)$, in each case.(i)![](https://cdn.mathpix.com/cropped/2025_10_30_83c6c2e6572790483bf8g-01.jpg?height=412&width=438&top_left_y=976&top_left_x=805)(ii)![](https://cdn.mathpix.com/cropped/2025_10_30_83c6c2e6572790483bf8g-01.jpg?height=418&width=452&top_left_y=1496&top_left_x=796)(iii)

Accepted Answer

(i) The number of zeroes is 0 as the graph does not cut the $x$-axis at any point.(ii) The number of zeroes is 1 as the graph intersects the $x$-axis at only 1 point.(iii) The number of zeroes is 3 as the graph intersects the $x$-axis at 3 points.(iv) The number of zeroes is 2 as the graph intersects the $x$-axis at 2 points.(v)The number of zeroes is 4 as the graph intersects the $x$-axis at 4 points.(vi) The number of zeroes is 3 as the graph intersects the $x$-axis at 3 points.\section*{Mathematics}(Chapter - 2) (Polynomials)(Class X)\section*{Exercise 2.2}

Question 2

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.(i) $x^{2}-2 x-8$(ii) $4 s^{2}-4 s+1$(iii) $6 x^{2}-3-7 x$(iv) $4 u^{2}+8 u$(v) $t^{2}-15$(vi) $3 x^{2}-x-4$

Accepted Answer

(i) $\quad x^{2}-2 x-8=(x-4)(x+2)$The value of $x^{2}-2 x-8$ is zero when $x-4=0$ or $x+2=0$, i.e., when $x =4$ or $x=-2$Therefore, the zeroes of $x^{2}-2 x-8$ are 4 and -2 .Sum of zeroes $=4-2=2=\frac{-(-2)}{1}=\frac{-(	ext { Coefficient of } x)}{	ext { Coefficient of } x^{2}}$Product of zeroes $=4 	imes(-2)=-8=\frac{(-8)}{1}=\frac{	ext { Constant term }}{	ext { Coefficient of } x^{2}}$(ii) $\quad 4 s^{2}-4 s+1=(2 s-1)^{2}$The value of $4 s^{2}-4 s+1$ is zero when $2 s-1=0$, i.e., $s=\frac{1}{2}$ Therefore, the zeroes of $4 s^{2}-4 s+1$ are $1 / 2$ and $1 / 2$.Sum of zeroes $=\frac{1}{2}+\frac{1}{2}=1=\frac{-(-4)}{4}=\frac{-(	ext { Coefficient of } s)}{\left(	ext { Coefficient of } s^{2}ight)}$Product of zeroes $=\frac{1}{2} 	imes \frac{1}{2}=\frac{1}{4}=\frac{	ext { Constant term }}{	ext { Coefficient of } s^{2}}$(iii) $6 x^{2}-3-7 x=6 x^{2}-7 x-3=(3 x+1)(2 x-3)$The value of $6 x^{2}-3-7 x$ is zero when $3 x+1=0$ or $2 x-3=0$, i.e.,$x=\frac{-1}{3} \quad$ or $\quad x=\frac{3}{2}$Therefore, the zeroes of $6 x^{2}-3-7 x$ are $\frac{-1}{3}$ and $\frac{3}{2}$.Sum of zeroes $=\frac{-1}{3}+\frac{3}{2}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-(	ext { Coefficient of } x)}{	ext { Coefficient of } x^{2}}$Product of zeroes $=\frac{-1}{3} 	imes \frac{3}{2}=\frac{-1}{2}=\frac{-3}{6}=\frac{	ext { Constant term }}{	ext { Coefficient of } x^{2}}$(iv) $4 u^{2}+8 u=4 u^{2}+8 u+0$$=4 u(u+2)$The value of $4 u^{2}+8 u$ is zero when $4 u=0$ or $u+2=0$, i.e., $u=0$ or $u=-2$Therefore, the zeroes of $4 u^{2}+8 u$ are 0 and -2 .Sum of zeroes $=0+(-2)=-2=\frac{-(8)}{4}=\frac{-(	ext { Coefficient of } u)}{	ext { Coefficient of } u^{2}}$Product of zeroes $=0 	imes(-2)=0=\frac{0}{4}=\frac{	ext { Constant term }}{	ext { Coefficient of } u^{2}}$(v) $t^{2}-15$$$\begin{aligned}& =t^{2}-0 t-15 \& =(t-\sqrt{15})(t+\sqrt{15})\end{aligned}$$The value of $t^{2}-15$ is zero when $t-\sqrt{15}=0$ or $t+\sqrt{15}=0$, i.e., when $t=\sqrt{15}$ or $t=-\sqrt{15}$Therefore, the zeroes of $t^{2}

Question 3

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.(i) $\frac{1}{4},-1$(ii) $\quad \sqrt{2}, \frac{1}{3}$(iii) $0, \sqrt{5}$(iv) 1,1(v) $-\frac{1}{4}, \frac{1}{4}$(vi) 4,1Answer 2:(i) $\frac{1}{4},-1$Let the polynomial be $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ and its zeroes be $\alpha$ and $\beta$.$$\begin{aligned}& \alpha+\beta=\frac{1}{4}

Accepted Answer

See solution.

Question 4

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following:(i) $p(x)=x^{3}-3 x^{2}+5 x-3, \quad g(x)=x^{2}-2$(ii) $p(x)=x^{4}-3 x^{2}+4 x+5, \quad g(x)=x^{2}+1-x$(iii) $p(x)=x^{4}-5 x+6, \quad g(x)=2-x^{2}$

Accepted Answer

(i) $\quad p(x)=x^{3}-3 x^{2}+5 x-3$$$q(x)=x^{2}-2$$$$\begin{array}{r}x ^ { 2 } - 2 \longdiv { \begin{array} { l r } { x ^ { 3 } - 3 x ^ { 2 } + 5 x - 3 } \{ x ^ { 3 } + 2 x } \{ - \quad + }\end{array} } \\hline \begin{array}{r}-3 x^{2}+7 x-3 \-3 x^{2}+6 \+\quad-\end{array} \\hline \begin{array}{r}7 x-9 \+\end{array}\end{array}$$Quotient $=x-3$Remainder $=7 x-9$(ii) $p(x)=x^{4}-3 x^{2}+4 x+5=x^{4}+0 x^{3}-3 x^{2}+4 x+5$$$q(x)=x^{2}+1-x=x^{2}-x+1$$$$\begin{aligned}& x ^ { 2 } - x + 1 \longdiv { x ^ { 4 } + 0 . x ^ { 3 } - 3 x ^ { 2 } + 4 x + 5 } x^{4}-x^{3}+x^{2} \&-\quad+\quad- x^{3}-4 x^{2}+4 x+5 \& x^{3}-x^{2}+x-\quad+\quad- \& \hline \begin{array}{r}-3 x^{2}+3 x+5 \-3 x^{2}+3 x-3 \+\quad+\end{array} \& \hline\end{aligned}$$Quotient $=x^{2}+x-3$Remainder $=8$(iii) $\quad p(x)=x^{4}-5 x+6=x^{4}+0 x^{2}-5 x+6$$$q(x)=2-x^{2}=-x^{2}+2$$$$\begin{aligned}& - x ^ { 2 } + 2 \longdiv { - x ^ { 2 } - 2 } \& \begin{array}{l}x^{4}+0 x^{2}-5 x+6 \x^{4}-2 x^{2} \\frac{-+}{2 x^{2}-5 x+6} \2 x^{2}-4 \\frac{-\quad+}{-5 x+10}\end{array}\end{aligned}$$Quotient $=-x^{2}-2$Remainder $=-5 x+10$

Chapter 2

Exercise 2.1

Exercise 2.2

Exercise 2.3