NCERT Solutions Class 10 Maths Chapter 2

Q.1 The graphs of y = p(x) are given in figures below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Ans.

(i) The number of zeroes is 0 as the given graph does not intersect the x-axis.
(ii) The number of zeroes is 1 as the given graph intersects the x-axis at 1 point only.
(iii) The number of zeroes is 3 as the given graph intersects the x-axis at 3 points only.
(iv) The number of zeroes is 2 as the given graph intersects the x-axis at 2 points only.
(v) The number of zeroes is 4 as the given graph intersects the x-axis at 4 points only.
(vi) The number of zeroes is 3 as the given graph intersects the x-axis at 3 points only.

Q.2

Ans.

$\begin{array}{l}\text{(i)}\\ {\mathrm{x}}^2-2\mathrm{x}-8={\mathrm{x}}^2-4\mathrm{x}+2\mathrm{x}-8\\ =\mathrm{x}(\mathrm{x}-4)+2(\mathrm{x}-4)\\ =(\mathrm{x}+2)(\mathrm{x}-4)\\ \text{So the value of}{\mathrm{x}}^2-2\mathrm{x}-8\text{is zero when}\mathrm{x}=-2\text{or}\mathrm{x}=4.\\ \text{Therefore, the zeroes of}{\mathrm{x}}^2-2\mathrm{x}-8\text{are}-2\text{and 4.}\\ \text{Now,}\\ \text{Sum of zeroes =}-2+4=2=\frac{-(-2)}{1}=\frac{-(\text{Coefficient of x)}}{{\text{Coefficient of x}}^2}\\ \text{Product of zeroes =}-2\times 4=\frac{-8}{1}=\frac{\text{Constant term}}{{\text{Coefficient of s}}^2}\\ \text{(ii)}\\ 4{\mathrm{s}}^2-4\mathrm{s}+1={\left(2\mathrm{s}-1\right)}^2\\ \text{So the value of}4{\mathrm{s}}^2-4\mathrm{s}+1\text{is zero when}\mathrm{x}=\frac{1}{2}.\\ \text{Therefore, the zeroes of}4{\mathrm{s}}^2-4\mathrm{s}+1\text{are}\frac{1}{2}\text{and}\frac{1}{2}\text{.}\\ \text{Now,}\\ \text{Sum of zeroes =}\frac{1}{2}+\frac{1}{2}=1=\frac{-(-4)}{4}=\frac{-(\text{Coefficient of s)}}{{\text{Coefficient of s}}^2}\\ \text{Product of zeroes =}\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}=\frac{\text{Constant term}}{{\text{Coefficient of s}}^2}\\ \text{(iii)}\\ 6{\mathrm{x}}^2-3-7\mathrm{x}=6{\mathrm{x}}^2-9\mathrm{x}+2\mathrm{x}-3\\ =3\mathrm{x}(2\mathrm{x}-3)+1(2\mathrm{x}-3)\\ =(3\mathrm{x}+1)(2\mathrm{x}-3)\\ \text{So the value of}6{\mathrm{x}}^2-3-7\mathrm{x}\text{is zero when}\mathrm{x}=\frac{-1}{3}\text{or}\mathrm{x}=\frac{3}{2}.\\ \text{Therefore, the zeroes of}6{\mathrm{x}}^2-3-7\mathrm{x}\text{are}\frac{-1}{3}\text{and}\frac{3}{2}\text{.}\\ \text{Now,}\\ \text{Sum of zeroes =}\frac{-1}{3}+\frac{3}{2}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-(\text{Coefficient of x)}}{{\text{Coefficient of x}}^2}\\ \text{Product of zeroes =}\frac{-1}{3}\times \frac{3}{2}=\frac{-1}{2}=\frac{\text{Constant term}}{{\text{Coefficient of x}}^2}\\ \text{(iv)}\\ 4{\mathrm{u}}^2+8\mathrm{u}=4\mathrm{u}(\mathrm{u}+2)\\ \text{So the value of}4\mathrm{u}(\mathrm{u}+2)\text{is zero when}\mathrm{u}=0\text{or}\mathrm{x}=-2.\\ \text{Therefore, the zeroes of}4\mathrm{u}(\mathrm{u}+2)\text{are}0\text{and}-2.\\ \text{Now,}\\ \text{Sum of zeroes =}0-2=\frac{-8}{4}=\frac{-(\text{Coefficient of u)}}{{\text{Coefficient of u}}^2}\\ \text{Product of zeroes =}0\times (-2)=0=\frac{0}{4}=\frac{\text{Constant term}}{{\text{Coefficient of u}}^2}\\ \text{(v)}\\ {\mathrm{t}}^2-15={\mathrm{t}}^2-{\left(\sqrt{15}\right)}^2=\left(\mathrm{t}-\sqrt{15}\right)\left(\mathrm{t}+\sqrt{15}\right)\\ \text{So the value of}{\mathrm{t}}^2-15\text{is zero when}\mathrm{t}=\sqrt{15}\text{or}\mathrm{x}=-\sqrt{15}.\\ \text{Therefore, the zeroes of}{\mathrm{t}}^2-15\text{are}\sqrt{15}\text{and}-\sqrt{15}\text{.}\\ \text{Now,}\\ \text{Sum of zeroes =}\sqrt{15}-\sqrt{15}=0=\frac{0}{1}=\frac{-(\text{Coefficient of t)}}{{\text{Coefficient of t}}^2}\\ \text{Product of zeroes =}\sqrt{15}\times (-\sqrt{15})=-15=\frac{-15}{1}=\frac{\text{Constant term}}{{\text{Coefficient of t}}^2}\\ \text{(vi)}\\ 3{\mathrm{x}}^2-\mathrm{x}-4=3{\mathrm{x}}^2-4\mathrm{x}+3\mathrm{x}-4\\ =\mathrm{x}(3\mathrm{x}-4)+1(3\mathrm{x}-4)=(3\mathrm{x}-4)(\mathrm{x}+1)\\ \text{So the value of}3{\mathrm{x}}^2-\mathrm{x}-4\text{is zero when}\mathrm{x}=\frac{4}{3}\text{or}\mathrm{x}=-1.\\ \text{Therefore, the zeroes of}3{\mathrm{x}}^2-\mathrm{x}-4\text{are}\frac{4}{3}\text{and}-1.\\ \text{Now,}\\ \text{Sum of zeroes =}\frac{4}{3}-1=\frac{1}{3}=\frac{-(-1)}{3}=\frac{-(\text{Coefficient of x)}}{{\text{Coefficient of x}}^2}\\ \text{Product of zeroes =}\frac{4}{3}\times (-1)=\frac{-4}{3}=\frac{\text{Constant term}}{{\text{Coefficient of x}}^2}\end{array}$

Q.3

Ans.

$\begin{array}{l}\text{(i)}\frac{1}{4},-1\\ {\text{Let the quadratic polynomial be ax}}^2+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=\frac{1}{4}=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=-1=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{and thus}\\ \mathrm{a}=4,\mathrm{b}=-1,\mathrm{c}=-4.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is 4x}}^2-\mathrm{x}-4.\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{2},\frac{1}{3}\\ {\text{Let the quadratic polynomial be ax}}^2+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=\sqrt{2}=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=\frac{1}{3}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{Now,}\\ \mathrm{\alpha }+\mathrm{\beta }=\sqrt{2}=\frac{3\sqrt{2}}{3}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \text{and}\mathrm{\alpha \beta }=\frac{1}{3}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\mathrm{a}=3,\mathrm{b}=-3\sqrt{2},\mathrm{c}=1.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is 3x}}^2-3\sqrt{2}\mathrm{x}+1.\\ \text{(iii) 0,}\sqrt{5}\\ {\text{Let the quadratic polynomial be ax}}^2+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=0=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=\sqrt{5}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\mathrm{a}=1,\mathrm{b}=0,\mathrm{c}=\sqrt{5}.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is x}}^2+\sqrt{5}.\\ \text{(iv) 1, 1}\\ {\text{Let the quadratic polynomial be ax}}^2+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=1=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=1=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\mathrm{a}=1,\mathrm{b}=-1,\mathrm{c}=1.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is x}}^2-\mathrm{x}+1.\\ \\ \text{(v)}\frac{-1}{4},\frac{1}{4}\\ {\text{Let the quadratic polynomial be ax}}^2+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=\frac{-1}{4}=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=\frac{1}{4}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\mathrm{a}=4,\mathrm{b}=1,\mathrm{c}=1.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is 4x}}^2+\mathrm{x}+1.\\ \text{(vi) 4, 1}\\ {\text{Let the quadratic polynomial be ax}}^2+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=4=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=1=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\mathrm{a}=1,\mathrm{b}=-4,\mathrm{c}=1.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is x}}^2-4\mathrm{x}+1.\end{array}$

Q.4

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2+5\mathrm{x}-3,\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^2-2\\ \\ {\mathrm{x}}^2-2\begin{array}{c}\mathrm{x}-3\\ \overline{){\mathrm{x}}^3-3{\mathrm{x}}^2+5\mathrm{x}-3}\end{array}\\ \underset{¯}{\begin{array}{l}{\mathrm{x}}^3-2\mathrm{x}\\ -+\end{array}}\\ -3{\mathrm{x}}^2+7\mathrm{x}-3\\ -3{\mathrm{x}}^2+6\\ \frac{+-}{7\mathrm{x}-9}\\ \text{Quotient}=\mathrm{x}-3\\ \text{Remainder}=7\mathrm{x}-9\\ \text{(ii)}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^4-3{\mathrm{x}}^2+4\mathrm{x}+5,\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^2+1-\mathrm{x}\\ \\ {\mathrm{x}}^2+1-\mathrm{x}\begin{array}{c}{\mathrm{x}}^2+\mathrm{x}-3\\ \overline{){\mathrm{x}}^4-3{\mathrm{x}}^2+4\mathrm{x}+5}\end{array}\\ \underset{¯}{\begin{array}{l}{\mathrm{x}}^4+{\mathrm{x}}^2-{\mathrm{x}}^3\\ --+\end{array}}\\ {\mathrm{x}}^3-4{\mathrm{x}}^2+4\mathrm{x}+5\\ \underset{¯}{\begin{array}{l}{\mathrm{x}}^3-{\mathrm{x}}^2+\mathrm{x}\\ -+-\end{array}}\\ -3{\mathrm{x}}^2+3\mathrm{x}+5\\ \underset{¯}{\begin{array}{l}-3{\mathrm{x}}^2+3\mathrm{x}-3\\ +-+\end{array}}\\ 8\\ \text{Quotient}={\mathrm{x}}^2+\mathrm{x}-3\\ \text{Remainder}=8\\ \text{(iii)}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^4-5\mathrm{x}+6,\mathrm{g}\left(\mathrm{x}\right)=2-{\mathrm{x}}^2\\ \\ 2-{\mathrm{x}}^2\begin{array}{c}-{\mathrm{x}}^2-2\\ \overline{){\mathrm{x}}^4-5\mathrm{x}+6}\end{array}\\ \underset{¯}{\begin{array}{l}{\mathrm{x}}^4-2{\mathrm{x}}^2\\ -+\end{array}}\\ 2{\mathrm{x}}^2-5\mathrm{x}+6\\ \underset{¯}{\begin{array}{l}2{\mathrm{x}}^2-4\\ -+\end{array}}\\ -5\mathrm{x}+10\\ \text{Quotient}=-{\mathrm{x}}^2-2\\ \text{Remainder}=-5\mathrm{x}+10\end{array}$

Q.5

Ans.

$\begin{array}{l}\text{(i)\hspace{0.17em}\hspace{0.17em}}{\mathrm{t}}^2-3,\text{2}{\mathrm{t}}^4+3{\mathrm{t}}^3-2{\mathrm{t}}^2-9\mathrm{t}-12\\ \\ {\mathrm{t}}^2-3\begin{array}{c}2{\mathrm{t}}^2+3\mathrm{t}+4\\ \overline{)\text{2}{\mathrm{t}}^4+3{\mathrm{t}}^3-2{\mathrm{t}}^2-9\mathrm{t}-12}\end{array}\\ \underset{¯}{\begin{array}{l}2{\mathrm{t}}^4-6{\mathrm{t}}^2\\ -+\end{array}}\\ 3{\mathrm{t}}^3+4{\mathrm{t}}^2-9\mathrm{t}-12\\ \underset{¯}{\begin{array}{l}3{\mathrm{t}}^3-9\mathrm{t}\\ -+\end{array}}\\ 4{\mathrm{t}}^2-12\\ \underset{¯}{\begin{array}{l}4{\mathrm{t}}^2-12\\ -+\end{array}}\\ 0\\ \text{Remainder = 0}\\ \text{So,}{\mathrm{t}}^2-3\text{is a factor of 2}{\mathrm{t}}^4+3{\mathrm{t}}^3-2{\mathrm{t}}^2-9\mathrm{t}-12.\\ \\ \text{(ii)}{\mathrm{x}}^2+3\mathrm{x}+1,\text{3}{\mathrm{x}}^4+5{\mathrm{x}}^3-7{\mathrm{x}}^2+2\mathrm{x}+2\\ \\ {\mathrm{x}}^2+3\mathrm{x}+1\begin{array}{c}3{\mathrm{x}}^2-4\mathrm{x}+2\\ \overline{)\text{3}{\mathrm{x}}^4+5{\mathrm{x}}^3-7{\mathrm{x}}^2+2\mathrm{x}+2}\end{array}\\ \underset{¯}{\begin{array}{l}3{\mathrm{x}}^4+9{\mathrm{x}}^3+3{\mathrm{x}}^2\\ ---\end{array}}\\ -4{\mathrm{x}}^3-10{\mathrm{x}}^2+2\mathrm{x}+2\\ \underset{¯}{\begin{array}{l}-4{\mathrm{x}}^3-12{\mathrm{x}}^2-4\mathrm{x}\\ +++\end{array}}\\ 2{\mathrm{x}}^2+6\mathrm{x}+2\\ \underset{¯}{\begin{array}{l}2{\mathrm{x}}^2+6\mathrm{x}+2\\ ---\end{array}}\\ 0\\ \text{Remainder = 0}\\ \text{So,}{\mathrm{x}}^2+3\mathrm{x}+1\text{is a factor of 3}{\mathrm{x}}^4+5{\mathrm{x}}^3-7{\mathrm{x}}^2+2\mathrm{x}+2.\\ \\ \text{(iii)}{\mathrm{x}}^3-3\mathrm{x}+1,{\text{x}}^5-4{\mathrm{x}}^3+{\mathrm{x}}^2+3\mathrm{x}+1\\ \\ {\mathrm{x}}^3-3\mathrm{x}+1\begin{array}{c}{\mathrm{x}}^2-1\\ \overline{){\text{x}}^5-4{\mathrm{x}}^3+{\mathrm{x}}^2+3\mathrm{x}+1}\end{array}\\ \underset{¯}{\begin{array}{l}{\mathrm{x}}^5-3{\mathrm{x}}^3+{\mathrm{x}}^2\\ -+-\end{array}}\\ -{\mathrm{x}}^3+3\mathrm{x}+1\\ \underset{¯}{\begin{array}{l}-{\mathrm{x}}^3+3\mathrm{x}-1\\ +-+\end{array}}\\ 2\\ \text{Remainder = 2}\\ \text{So,}{\mathrm{x}}^3-3\mathrm{x}+1{\text{is not a factor of x}}^5-4{\mathrm{x}}^3+{\mathrm{x}}^2+3\mathrm{x}+1.\end{array}$

Q.6

Ans.

$\begin{array}{l}{\text{p(x)=3x}}^{\text{4}}{\text{+6x}}^{\text{3}}-{\text{2x}}^{\text{2}}-\text{10x-5}\\ \text{The two zeroes of p(x) are}\sqrt{\frac{\text{5}}{\text{3}}}\text{and-}\sqrt{\frac{\text{5}}{\text{3}}}\text{.}\\ \text{Therefore,}\left(\text{x}-\sqrt{\frac{\text{5}}{\text{3}}}\right)\text{and}\left(\text{x+}\sqrt{\frac{\text{5}}{\text{3}}}\right)\text{are factors of p(x).}\\ \text{Also,}\\ \left(\text{x}-\sqrt{\frac{\text{5}}{\text{3}}}\right)\left(\text{x+}\sqrt{\frac{\text{5}}{\text{3}}}\right){\text{\hspace{0.33em}=\hspace{0.33em}x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\\ {\text{and so x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\text{is a factor of p(x).}\\ \text{Now,}\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3x}}^{\text{2}}\text{+6x+3}\\ {\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\overline{){\text{3x}}^{\text{4}}{\text{+6x}}^{\text{3}}-{\text{2x}}^{\text{2}}-\text{10x}-\text{5}}\\ \underset{¯}{\begin{array}{l}{\text{3x}}^{\text{4}}-{\text{5x}}^{\text{2}}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}6x}}^{\text{3}}{\text{+3x}}^{\text{2}}-\text{10x}-\text{5}\\ \underset{¯}{\begin{array}{l}{\text{6x}}^{\text{3}}-\text{10x}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3x}}^{\text{2}}-\text{5}\\ \underset{¯}{\begin{array}{l}{\text{3x}}^{\text{2}}-\text{5}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0}\\ {\text{3x}}^{\text{4}}{\text{+6x}}^{\text{3}}-{\text{2x}}^{\text{2}}-\text{10x}-\text{5=}\left({\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\right)\left({\text{3x}}^{\text{2}}\text{+6x+3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3}\left({\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\right)\left({\text{x}}^{\text{2}}\text{+2x+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3}\left({\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\right)\left(\text{x+1}\right)\text{(x+1)}\\ \text{Equating}\left({\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\right)\left(\text{x+1}\right)\text{(x+1) equal to zero, we get}\\ \text{the zeroes of the given polynomial.}\\ \text{Hence, the zeroes of the given polynomial are}\sqrt{\frac{\text{5}}{\text{3}}}\text{,\hspace{0.17em}}-\sqrt{\frac{\text{5}}{\text{3}}}\text{,\hspace{0.17em}\hspace{0.33em}1}\\ \text{and\hspace{0.33em}}-\text{1.}\end{array}$

Q.7

Ans.

$\begin{array}{l}\text{Dividend =}{\mathrm{x}}^3-3{\mathrm{x}}^2+\mathrm{x}+2\\ \text{Quotient =}\mathrm{x}-2\\ \text{Remainder =}-2\mathrm{x}+4\\ \text{Divisor = g(x) = ?}\\ \text{We know that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividend}=\text{Divisor}\times \text{Quotient}+\text{Remainder}\\ \therefore {\mathrm{x}}^3-3{\mathrm{x}}^2+\mathrm{x}+2=\text{g(x)}\times (\mathrm{x}-2)+(-2\mathrm{x}+4)\\ \Rightarrow \text{g(x)}\times (\mathrm{x}-2)={\mathrm{x}}^3-3{\mathrm{x}}^2+\mathrm{x}+2+2\mathrm{x}-4\\ \Rightarrow \text{g(x)}=\frac{({\mathrm{x}}^3-3{\mathrm{x}}^2+3\mathrm{x}-2)}{(\mathrm{x}-2)}\\ \\ \mathrm{x}-2\begin{array}{c}{\mathrm{x}}^2-\mathrm{x}+1\\ \overline{){\mathrm{x}}^3-3{\mathrm{x}}^2+3\mathrm{x}-2}\end{array}\\ \underset{¯}{\begin{array}{l}{\mathrm{x}}^3-2{\mathrm{x}}^2\\ -+\end{array}}\\ -{\mathrm{x}}^2+3\mathrm{x}-2\\ \underset{¯}{\begin{array}{l}-{\mathrm{x}}^2+2\mathrm{x}\\ +-\end{array}}\\ \mathrm{x}-2\\ \underset{¯}{\begin{array}{l}\mathrm{x}-2\\ -+\end{array}}\\ 0\\ \therefore \mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^2-\mathrm{x}+1\end{array}$

Q.8

Ans.

$\begin{array}{l}\text{(i) deg}\mathrm{p}\left(\mathrm{x}\right)=\text{deg}\mathrm{q}\left(\mathrm{x}\right)\\ \text{if divisor is constant then}\\ \text{Degree of quotient = degree of dividend}\\ \text{Let dividend}\mathrm{p}\left(\mathrm{x}\right)=2{\mathrm{x}}^2-2\mathrm{x}+14\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}divisor \hspace{0.17em}}\mathrm{g}\left(\mathrm{x}\right)=2\\ \text{Then, wer have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}quotient\hspace{0.17em}\hspace{0.17em}}\mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^2-\mathrm{x}+7\\ \text{and remainder\hspace{0.17em}\hspace{0.17em}}\mathrm{r}\left(\mathrm{x}\right)=0\\ \text{We find that}\\ \text{deg}\mathrm{p}\left(\mathrm{x}\right)=\text{deg}\mathrm{q}\left(\mathrm{x}\right)=2\\ \text{Let us check for division algorithm.}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividend}=\text{divisor}\times \text{quotient + remainder}\\ \text{or}2{\mathrm{x}}^2-2\mathrm{x}+14=2({\mathrm{x}}^2-\mathrm{x}+7)+0=2{\mathrm{x}}^2-2\mathrm{x}+14\\ \text{Thus, the division algorithm is satisfied.}\\ \text{(ii) deg}\mathrm{q}\left(\mathrm{x}\right)=\text{deg}\mathrm{r}\left(\mathrm{x}\right)\\ \\ \text{Let us divide}{\mathrm{x}}^3+\mathrm{x}\text{by}{\mathrm{x}}^2.\\ \text{Clearly we have,}\\ \text{dividend}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^3+\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}divisor \hspace{0.17em}}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}quotient\hspace{0.17em}\hspace{0.17em}}\mathrm{q}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{and remainder\hspace{0.17em}\hspace{0.17em}}\mathrm{r}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{Here,}\\ \text{deg}\mathrm{q}\left(\mathrm{x}\right)=\text{deg}\mathrm{r}\left(\mathrm{x}\right)=1\\ \text{Let us check for division algorithm.}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividend}=\text{divisor}\times \text{quotient + remainder}\\ \text{or}{\mathrm{x}}^3+\mathrm{x}={\mathrm{x}}^2\cdot \mathrm{x}+\mathrm{x}={\mathrm{x}}^3+\mathrm{x}\\ \text{Thus, the division algorithm is satisfied.}\\ \text{(iii) deg}\mathrm{r}\left(\mathrm{x}\right)=0\\ \\ \text{Degree of remainder will be zero if the remainder is constant.}\\ \text{Let us divide}{\mathrm{x}}^2+1\text{by}\mathrm{x}.\\ \text{Clearly we have,}\\ \text{dividend}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^2+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}divisor \hspace{0.17em}}\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}quotient\hspace{0.17em}\hspace{0.17em}}\mathrm{q}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{and remainder\hspace{0.17em}\hspace{0.17em}}\mathrm{r}\left(\mathrm{x}\right)=1\\ \text{Here,}\\ \text{deg}\mathrm{r}\left(\mathrm{x}\right)=0\\ \text{Let us check for division algorithm.}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividend}=\text{divisor}\times \text{quotient + remainder}\\ \text{or}{\mathrm{x}}^2+1=\mathrm{x}\cdot \mathrm{x}+1={\mathrm{x}}^2+1\\ \text{Thus, the division algorithm is satisfied.}\end{array}$

Q.9

Ans.

$\begin{array}{l}{\text{(i) 2x}}^3+{\mathrm{x}}^2-5\mathrm{x}+2;\frac{1}{2},1,-2\\ \text{Let}\mathrm{p}\left(\mathrm{x}\right)={\text{2x}}^3+{\mathrm{x}}^2-5\mathrm{x}+2\\ \text{Then}\\ \mathrm{p}\left(\frac{1}{2}\right)=\text{2}{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^2-5\left(\frac{1}{2}\right)+2\\ =\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=\frac{1}{2}+2-\frac{5}{2}=0\\ \\ \mathrm{p}\left(1\right)=\text{2}{\left(1\right)}^3+{\left(1\right)}^2-5\left(1\right)+2\\ =2+1-5+2=0\\ \mathrm{p}(-2)=\text{2}{(-2)}^3+{(-2)}^2-5(-2)+2\\ =-16+4+10+2=0\\ \\ \text{Therefore,}\frac{1}{2},1\text{and\hspace{0.17em}\hspace{0.17em}}-2\text{are zeroes of the given poynomial.}\\ \text{Comparing the given poynomial with}{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d},\text{we get}\\ \mathrm{a}=2,\mathrm{b}=1,\mathrm{c}=-5,\mathrm{d}=2\\ \text{Let the given roots are}\mathrm{\alpha },\mathrm{\beta }\text{and}\mathrm{\gamma }\text{. Then}\\ \mathrm{\alpha }\text{\hspace{0.17em}=\hspace{0.17em}}\frac{1}{2},\mathrm{\beta }=1,\mathrm{\gamma }=-2\\ \text{Now,}\\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=\frac{1}{2}+1-2=-\frac{1}{2}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=\frac{1}{2}\times 1+1\times (-2)+(-2)\times \frac{1}{2}=\frac{-5}{2}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{\alpha \beta \gamma }=\frac{1}{2}\times 1\times (-2)=-1=\frac{-2}{2}=\frac{-\mathrm{d}}{\mathrm{a}}\\ \\ \text{The relationship between roots and coefficients is verified.}\\ {\text{(ii) x}}^3-4{\mathrm{x}}^2+5\mathrm{x}-2;\text{2, 1, 1}\\ \text{Let}\mathrm{p}\left(\mathrm{x}\right)={\text{x}}^3-4{\mathrm{x}}^2+5\mathrm{x}-2\\ \text{Then}\\ \mathrm{p}\left(\text{2}\right)={\left(2\right)}^3-4{\left(2\right)}^2+5\left(2\right)-2\\ =8-16+10-2=0\\ \\ \mathrm{p}\left(1\right)={\left(1\right)}^3-4{\left(1\right)}^2+5\left(1\right)-2\\ =1-4+5-2=0\\ \\ \text{Therefore,}2,1\text{and\hspace{0.17em}\hspace{0.17em}}1\text{are zeroes of the given poynomial.}\\ \text{Comparing the given poynomial with}{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d},\text{we get}\\ \mathrm{a}=1,\mathrm{b}=-4,\mathrm{c}=5,\mathrm{d}=-2\\ \text{Let the given roots are}\mathrm{\alpha },\mathrm{\beta }\text{and}\mathrm{\gamma }\text{. Then}\\ \mathrm{\alpha }\text{\hspace{0.17em}=\hspace{0.17em}}2,\mathrm{\beta }=1,\mathrm{\gamma }=1\\ \text{Now,}\\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=2+1+1=4=-\frac{-4}{1}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=2\times 1+1\times 1+1\times 2=5=\frac{5}{1}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{\alpha \beta \gamma }=2\times 1\times 1=2=\frac{-(-2)}{1}=\frac{-\mathrm{d}}{\mathrm{a}}\\ \text{The relationship between roots and coefficients is verified.}\end{array}$

Q.10

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Q.11

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Q.12

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Q.13

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