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) Diag $\\$ (ii) Diag $\\$ (iii) Diag $\\$ (iv) Diag $\\$ (v) Diag $\\$ (vi) Diag $\\$

Solution :

(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. $\\$

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

Solution :

(iv) $4u^2 + 8u = 4u^2 + 8u + 0 = 4u(u+2) $ $\\$ The value of $4u^2 + 8u$ is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = -2 $\\$ Therefore, the zeroes of $4u^2 + 8u$ are 0 and -2. $\\$ Sum of zeroes = $0 + (-2) = -2 = \frac{-(8)}{4} = \frac{-(Coefficient \ of \ u)}{Coefficient \ of \ u^2}$ $\\$ Product of zeroes = $0 \times (-2) = 0 = \frac{0}{4} = \frac{Constant \ term}{Coefficient \ of \ u^2}$ $\\$

(i) $x^2 – 2x – 8 = (x-4)(x+2)$ $\\$ The value of $x^2 – 2x – 8$ is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2 $\\$ Therefore, the zeroes of $x^2 – 2x – 8$ Sum of zeroes = $4-2=2 = \frac{-(-2)}{1} = \frac{-(Coefficient \ of \ x)}{Coefficient \ of \ x^2}$ $\\$ Product of zeroes = $4 \times (-2) = -8 = \frac{(-8)}{1} = \frac{Constant \ term}{Coefficient \ of \ x^2}$ $\\$ (ii) $4s^2 – 4s + 1 = (2s-1)^2$ $\\$ The value of $4s^2 – 4s + 1$ is zero when 2s - 1 = 0, i.e., $s = \frac{1}{2}$ $\\$ Therefore, the zeroes of $4s^2 – 4s + 1$ are $\frac{1}{2}$ and $\frac{1}{2}$ $\\$ Sum of zeroes = $\frac{1}{2} + \frac{1}{2} = 1 = \frac{-(-4)}{4} = \frac{-(Coefficient \ of \ s)}{Coefficient \ of \ s^2}$ $\\$ Product of zeroes $= \frac{1}{2} \times \frac{1}{2}= \frac{1}{4} = \frac{Constant \ term }{Coefficient \ of \ s^2}$ $\\$ (iii) $6x^2 - 3 - 7x = 6x^2-7x-3 = (3x+1)(2x-3)$ $\\$ The value of $6x^2 - 3 - 7x$ s zero when 3x + 1 = 0 or 2x - 3= 0, i.e., x $= \frac{-1}{3}$ or x = $\frac{3}{2}$ $\\$ Therefore, the zeroes of $6x^2 - 3 - 7x$ are $\frac{-1}{3}$ and $\frac{3}{2}$ $\\$ Sum of zeroes $= \frac{-1}{3}+ \frac{3}{2} = \frac{7}{6}= \frac{-(-7)}{6} = \frac{-(Coefficient \ of \ x)}{Coefficient \ of \ x^2}$ $\\$ Product of zeroes = $\frac{-1}{3} \times \frac{3}{2} = \frac{-1}{2} = \frac{-3}{6} = \frac{Constant \ term}{Coefficient \ of \ x^2}$ $\\$ (iv) $4u^2 + 8u = 4u^2 + 8u + 0 = 4u(u+2) $ $\\$ The value of $4u^2 + 8u$ is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = -2 $\\$ Therefore, the zeroes of $4u^2 + 8u$ are 0 and -2. $\\$ Sum of zeroes = $0 + (-2) = -2 = \frac{-(8)}{4} = \frac{-(Coefficient \ of \ u)}{Coefficient \ of \ u^2}$ $\\$ Product of zeroes = $0 \times (-2) = 0 = \frac{0}{4} = \frac{Constant \ term}{Coefficient \ of \ u^2}$ $\\$

(ii) $4s^2 – 4s + 1 = (2s-1)^2$ $\\$ The value of $4s^2 – 4s + 1$ is zero when 2s - 1 = 0, i.e., $s = \frac{1}{2}$ $\\$ Therefore, the zeroes of $4s^2 – 4s + 1$ are $\frac{1}{2}$ and $\frac{1}{2}$ $\\$ Sum of zeroes = $\frac{1}{2} + \frac{1}{2} = 1 = \frac{-(-4)}{4} = \frac{-(Coefficient \ of \ s)}{Coefficient \ of \ s^2}$ $\\$ Product of zeroes $= \frac{1}{2} \times \frac{1}{2}= \frac{1}{4} = \frac{Constant \ term }{Coefficient \ of \ s^2}$ $\\$

(v) $t^2 - 15 = t^2 - 0t -15 = (t- \sqrt{15})(t + \sqrt{15})$ $\\$ 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 $t^2 - 15$ are $\sqrt{15}$ and $-\sqrt{15}$ $\\$ Sum of zeroes = $\sqrt{15} + (- \sqrt{15}) = 0 = \frac{-0}{1} = \frac{-(Coefficient \ of \ t)}{Coefficient \ of \ t^2}$ $\\$ Product of zeroes = $(\sqrt{15} (-\sqrt{15})) = -15 = \frac{15}{1} = \frac{Constant \ term}{Coefficient \ of \ x^2}$ $\\$ (vi) $3x^2 - x - 4 = (3x-4) = (x+1)$ $\\$ The value of $3x^2 - x - 4$ is zero when 3x - 4 = 0 or x + 1 = 0, i.e., when $x = \frac{4}{3}$ or x = -1 $\\$ Therefore, the zeroes of $3x^2 - x - 4$ are $\frac{4}{3}$ and -1 $\\$ Sum of zeroes = $\frac{4}{3} + (-1 ) = \frac{4-3}{3} = \frac{1}{3} = \frac{-(-1)}{3} = \frac{-(Coefficient \ of \ x)}{Coefficient \ of \ x^2}$ $\\$ Product of zeroes = $\frac{4}{3} \times (-1) = - \frac{4}{3} = \frac{-4}{3} = \frac{Constant \ term}{Coefficient \ of \ x^2}$ $\\$

Therefore, the zeroes of $x^2 – 2x – 8$ Sum of zeroes = $4-2=2 = \frac{-(-2)}{1} = \frac{-(Coefficient \ of \ x)}{Coefficient \ of \ x^2}$ $\\$

3   Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.$\\$ (i) $\frac{1}{4}, -1 \qquad $ (ii) $\sqrt{2}, \frac{1}{3} \qquad $ (iii) $0, \sqrt{5}$ $\\$ (iv) $1,1 \qquad$ (v) $-\frac{1}{4} , \frac{1}{4} \qquad$ (vi) $4,1$ $\\$

Solution :

(i) $\frac{1}{4}, -1$ $\\$ Let the polynomial be $ax^2 + bx+c$ , and its zeroes be $\alpha$ and $\beta$. $\\$ $\alpha + \beta = \frac{1}{4} = \frac{-b}{a}$ $\\$ $\alpha \beta = -1 = \frac{-4}{4} = \frac{c}{a}$ $\\$ If $a = 4 $ then $b = -1, c= -4$ $\\$ Therefore, the quadratic polynomial is $4x^2 - x - 4.$ $\\$ (ii) $\sqrt{2}, \frac{1}{3}$ $\\$ Let the polynomial be $ax^2 + bx+c$ , and its zeroes be $\alpha$ and $\beta$. $\\$ $\alpha + \beta = \sqrt{2} = \frac{3\sqrt{2}}{3} = \frac{-b }{a}$ $\\$ $\alpha \beta = \frac{1}{3} = \frac{c}{a}$ $\\$ If $a =3$ then $b = -3 \sqrt{2}, c=1$ $\\$ Therefore, the quadratic polynomial is $3x^2 - 3 \sqrt{2} x + 1$ $\\$ (iii) $0, \sqrt{5}$ $\\$ Let the polynomial be $ax^2 + bx+c$ , and its zeroes be $\alpha$ and $\beta$. $\\$ $\alpha + \beta = 0 = \frac{0}{1} = \frac{-b}{a}$ $\\$ $\alpha \times \beta = \sqrt{5} = \frac{\sqrt{5}}{1} = \frac{c}{a}$ $\\$ If a = 1 then b = 0, c = $\sqrt{5}$ $\\$ Therefore, the quadratic polynomial is $x^2 + \sqrt{5}$ $\\$ (iv) $1,1$ Let the polynomial be $ax^2 + bx+c$ , and its zeroes be $\alpha$ and $\beta$. $\\$ $\alpha + \beta = 1 = \frac{1}{1} = \frac{-b}{a}$ $\\$ $\alpha \times \beta = 1 = \frac{1}{1} = \frac{c}{a}$ $\\$ If a = 1 then b = -1, c = 1 $\\$ Therefore, the quadratic polynomial is $x^2 - x +1$ $\\$

(v) $-\frac{1}{4} , \frac{1}{4}$ Let the polynomial be $ax^2 + bx+c$ , and its zeroes be $\alpha$ and $\beta$. $\\$ $\alpha + \beta = \frac{-1}{4} = \frac{-b}{a} $ $\\$ $\alpha \times \beta = \frac{1}{4} = \frac{c}{a}$ $\\$ If a = 4, then b = 1, c = 1 $\\$ Therefore, the quadratic polynomial is $4x^2 + x + c$ $\\$ (vi) $4,1$ $\\$ Let the polynomial be $ax^2 + bx+c$ $\\$ $\alpha + \beta = 4 = \frac{4}{1} = \frac{-b}{a}$ $\\$ $\alpha \times \beta = 1 = \frac{1}{1} = \frac{c}{a}$ $\\$ If a = 1 then b = -4, c = 1 $\\$ Therefore, the quadratic polynomial is $x^2 - 4x +1.$ $\\$

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 - 3x^2 + 5x -3, \qquad g(x) = x^2 -2 $ $\\$ (ii) $p(x) = x^4 - 3x^2 + 4x + 5, \qquad g(x) = x^2 +1-x $ $\\$ (iii) $p(x) = x^4 - 5x + 6, \qquad g(x) = 2 - x^2 $ $\\$ $\\$

Solution :

(i) $p(x) = x^3 - 3x^2 + 5x -3$, $\\$ $q(x) = x^2 -2 $ $\\$ $$ \require{enclose} \begin{array}{rll} x-3 && \\[-3pt] x^2 -2 \enclose{longdiv}{x^3 - 3x^2 + 5x -3}\kern-.2ex \\[-3pt] {x^3 \qquad \ \ \quad - 2x \phantom{00}} && \\[-3pt] \underline{- \qquad \ \ \qquad + \phantom{00}} && \\[-3pt] -3x^2 + 7x -3 \phantom{0} && \\[-3pt] {\phantom{0} -3x^2 \qquad +6 \phantom{0}} && \\[-3pt] \underline{ \ \ + \quad \ \ \qquad - \phantom{00}} && \\[-3pt] \underline{\phantom{0} \qquad \ \ \ \ \ 7x-9 \ \ \ } && \\[-3pt] \phantom{00} \end{array} $$ Quotient = x - 3 $\\$ Remainder = 7x - 9 $\\$ $\\$ (ii) $p(x) = x^4 - 3x^2 + 4x + 5 = x^4 +0x^3 - 3x^2 + 4x + 5$ $\\$ $g(x) = x^2 +1-x = x^2 -x+1 $ $\\$ $$ \require{enclose} \begin{array}{rll} x^2 + x-3 && \\[-3pt] x^2 -x+1 \enclose{longdiv}{x^4 + 0x^3 - 3x^2 + 4x +5}\kern-.2ex \\[-3pt] {x^4 - x^3 + x^2 \qquad \ \ \quad \ \ \phantom{00}} && \\[-3pt] \underline{- \quad + \quad - \qquad \ \ \qquad \phantom{00}} && \\[-3pt] x^3 - 4x^2+4x+5 \phantom{0} && \\[-3pt] {\phantom{0} x^3 \ - \ x2 \ + x \qquad \phantom{0}} && \\[-3pt] \underline{ \ \ - \quad + \quad \ \ - \qquad \phantom{00}} && \\[-3pt] {\phantom{0} -3x^2 \ + 3x + 5 \phantom{0}} && \\[-3pt] {\phantom{0} -3x^2 \ + 3x - 3 \phantom{0}} && \\[-3pt] \underline{ \ \ + \quad - \quad \ \ + \phantom{00}} \\[-3pt] \underline{\phantom{0} \qquad \ \ \ \qquad \ \ \ \ 8 \ \ \ } && \\[-3pt] \phantom{00} \end{array} $$ Quotient = $x^2 + x - 3$ Remainder = 8

(iii) $p(x) = x^4 - 5x + 6 = x^4 + 0x^2 - 5x + 6 $ $\\$ $g(x) = 2 - x^2= - x^2 + 2 $ $\\$ $$ \require{enclose} \begin{array}{rll} -x^2 - 2 && \\[-3pt] -x^2 + 2 \enclose{longdiv}{x^4 + 0x^2 - 5x + 6}\kern-.2ex \\[-3pt] {x^4 - 2x^2 \qquad \quad \phantom{00}} && \\[-3pt] \underline{- \quad + \quad \ \ \qquad \phantom{00}} && \\[-3pt] 2x^2 - 5x +6 \phantom{0} && \\[-3pt] {\phantom{0} 2x^2 \qquad -4 \phantom{0}} && \\[-3pt] \underline{ \ \ - \quad \ \ \qquad + \phantom{00}} && \\[-3pt] \underline{\phantom{0} \qquad -5x+10 \ \ \ } && \\[-3pt] \phantom{00} \end{array} $$ Quotient = -x 2 - 2 $\\$ Remainder = -5x +10 $\\$

5   Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: $\\$ (i) $t^2 - 3, 2t^4 + 3t^3 - 2t^2 - 9t - 12$ $\\$ (ii) $x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2$ $\\$ (iii) $x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1$ $\\$

Solution :

(i) $t^2 - 3, 2t^4 + 3t^3 - 2t^2 - 9t - 12$ $\\$ $t^2 -3 = t^2 + 0t-3$ $\\$ $$ \require{enclose} \begin{array}{rll} 2t^2 + 3t + 4 && \\[-3pt] t^2 + 0t - 3\enclose{longdiv}{2t^4 + 3t^3 - 2t^2 - 9t - 12}\kern-.2ex \\[-3pt] {2t^4 + 0t^3 - 6t^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{- \quad - \quad + \qquad \qquad \ \ \phantom{00}} && \\[-3pt] 3t^3 + 4t^2 - 9t - 12 \phantom{0} && \\[-3pt] {\phantom{0} 3t^3 + 0t^2 - 9t \qquad \phantom{0}} && \\[-3pt] \underline{- \quad - \quad + \qquad \quad \phantom{00}} && \\[-3pt] 4t^2 + 0t - 12 \phantom{0} && \\[-3pt] {\phantom{0} 4t^2 + 0t - 12 \phantom{0}} && \\[-3pt] \underline{- \quad - \quad + \phantom{00}} && \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ 0 \ \ } && \\[-3pt] \phantom{00} \end{array} $$ $\\$ Since the remainder is 0, $\\$ Hence, $t^2 -3 $ is a factor of $ 2t^4 + 3t^3 - 2t^2 - 9t - 12$

(iii) $x^3 - 3x + 1, x^5 - 4x^3 + x^2 + 3x + 1$ $\\$ $$ \require{enclose} \begin{array}{rll} x^2 - 1 && \\[-3pt] x^3 - 3x + 1 \enclose{longdiv}{ x^5 - 4x^3 + x^2 + 3x + 1}\kern-.2ex \\[-3pt] {x^5 - 3x^3 + x^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{- \quad + \quad - \qquad \qquad \ \ \phantom{00}} && \\[-3pt] -x^3 \quad + 3x + 1 \phantom{0} && \\[-3pt] {\phantom{0} -x^3 \quad + 3x - 1 \phantom{0}} && \\[-3pt] \underline{+ \qquad - \quad \ + \phantom{00}} && \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ \ \ 2 \ \ } && \\[-3pt] \phantom{00} \end{array} $$ $\\$ Since the remainder $\neq$ 0, $\\$ Hence, $ x^3 - 3x + 1$ is not a factor of $x^5 - 4x^3 + x^2 + 3x + 1$ $\\$

(ii) $x^2 + 3x + 1, 3x^4 + 5x^3 - 7x^2 + 2x + 2$ $\\$ $$ \require{enclose} \begin{array}{rll} 3x^2 - 4x + 2 && \\[-3pt] x^2 + 3x + 1 \enclose{longdiv}{3x^4 + 5x^3 - 7x^2 + 2x + 2}\kern-.2ex \\[-3pt] {3x^4 + 5x^3 - 7x^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{- \quad - \quad - \qquad \qquad \ \ \phantom{00}} && \\[-3pt] -4x^3 - 10x^2 + 2x + 2 \phantom{0} && \\[-3pt] {\phantom{0} -4x^3 - 12x^2 - 4x \quad \ \ \phantom{0}} && \\[-3pt] \underline{+ \qquad + \qquad + \qquad \phantom{00}} && \\[-3pt] 2x^2 + 6x + 2 \phantom{0} && \\[-3pt] \underline{\phantom{0} 2x^2 + 6x + 2 \phantom{0}} && \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ 0 \ \ } && \\[-3pt] \phantom{00} \end{array} $$ $\\$ Since the remainder is 0, $\\$ Hence, $x^2 + 3x +1$ is a factor of $3x^4 + 5x^3 - 7x^2 + 2x + 2.$

6   Obtain all other zeroes of $3x^4 + 6x^3 - 2x^2 -10x -5$ if two of its zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ $\\$

Solution :

p(x) = $3x^4 + 6x^3 - 2x^2 -10x -5$ $\\$ Since the two zeroes are $\sqrt{\frac{5}{3}}$ and $ - \sqrt{\frac{5}{3}}$, $\\$ $\therefore \Bigg( x - \sqrt{\frac{5}{3}} \Bigg) \Bigg( x + \sqrt{\frac{5}{3}} \Bigg) = \Bigg( x^2 - \frac{5}{3} \Bigg)$ is a factor of $\\$ Therefore, we divide the given polynomial by $x^2 - \frac{5}{3}$ $\\$ $$ \require{enclose} \begin{array}{rll} 3x^2 + 6x + 3 && \\[-3pt] x^2 + 0x - \frac{5}{3} \enclose{longdiv}{3x^4 + 6x^3 - 2x^2 - 10x - 5}\kern-.2ex \\[-3pt] {3x^4 + 0x^3 - 5x^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{- \quad - \quad + \qquad \qquad \ \ \phantom{00}} && \\[-3pt] 6x^3 + 3x^2 - 10x - 5 \phantom{0} && \\[-3pt] {\phantom{0} 6x^3 + 0x^2 - 10x \quad \ \ \phantom{0}} && \\[-3pt] \underline{- \qquad - \qquad + \qquad \phantom{00}} && \\[-3pt] 3x^2 + 0x - 5 \phantom{0} && \\[-3pt] {\phantom{0} 3x^2 + 0x - 5 \phantom{0}} && \\[-3pt] \underline{- \quad - \quad + \phantom{00}} && \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ 0 \ \ } && \\[-3pt] \phantom{00} \end{array} $$ $3x^4 + 6x^3 - 2x^2 - 10x - 5 = \bigg( x^2 - \frac{5}{3} \bigg) \big( 3x^2 + 6x + 3 \big)$ $\\$ $\qquad = 3 \bigg( x^2 - \frac{5}{3} \bigg) \big( x^2 + 2x + 1 \big)$ $\\$ We factorize $x^2 + 2x + 1$ $\\$ $ = (x+1)^2$ $\\$ Therefore, its zero is given by x + 1 = 0 $\\$ x = -1 $\\$ As it has the term $(x+1)^2$ $\\$ , therefore, there will be 2 zeroes at x = -1. $\\$ Hence, the zeroes of the given polynomial are $\sqrt{\frac{5}{3}}, - \sqrt{\frac{5}{3}}$, -1 and -1. $\\$

7   On dividing $x^3 – 3x^2 + x + 2$ by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x). $\\$

Solution :

p(x) = $x^3 – 3x^2 + x + 2$ $\\$ g(x) = ? (Divisor) $\\$ Quotient = (x - 2) $\\$ Remainder = (- 2x + 4) $\\$ Dividend = Divisor $\times$ Quotient + Remainder $\\$ $x^3 – 3x^2 + x + 2 = g(x) \times (x-2) + (-2x + 4) $ $\\$ $x^3 – 3x^2 + x + 2 + 2x -4 = g(x) (x -2)$ $\\$ $x^3 - 3x^2 + 3x -2 = g(x) (x -2)$ $\\$ g(x) is the quotient when we divide $(x^3 - 3x^2 + 3x -2)$ by $(x-2)$ $\\$ $$ \require{enclose} \begin{array}{rll} x^2 - x + 1 && \\[-3pt] x-2 \enclose{longdiv}{x^3 – 3x^2 + 3x - 2}\kern-.2ex \\[-3pt] {x^3 - 2x^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{ - \quad + \qquad \qquad \ \ \phantom{00}} && \\[-3pt] -x^2 + 3x - 2 \phantom{0} && \\[-3pt] {\phantom{0} -x^2 + 2x \quad \ \ \phantom{0}} && \\[-3pt] \underline{ + \quad - \qquad \phantom{00}} && \\[-3pt] x - 2 \phantom{0} && \\[-3pt] {\phantom{0} x - 2 \phantom{0}} && \\[-3pt] \underline{ - \quad + \phantom{00}} && \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ 0 \ \ } && \\[-3pt] \phantom{00} \end{array} $$ $\therefore g(x) = (x^2 - x + 1)$ $\\$

8   Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and $\\$ (i) deg p(x) = deg q(x) $\qquad$ (ii) deg q(x) = deg r(x) $\qquad$ (iii) deg r(x) = 0 $\\$

Solution :

According to the division algorithm, if p(x) and g(x) are two polynomials with $\\$ g(x) $\neq$ 0, then we can find polynomials q(x) and r(x) such that $\\$ p(x) = g(x) $\times$ q(x) + r(x), $\\$ where r(x) = 0 or degree of r(x) $<$ degree of g(x) $\\$ Degree of a polynomial is the highest power of the variable in the polynomial. $\\$ (i) deg p(x) = deg q(x) Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant). $\\$ Let us assume the division of $6x^2 + 2x + 2$ by 2. $\\$ Here, p(x) = $\\$ g(x) = 2 $\\$ q(x) = $3x^2 + x + 1$ and r(x) = 0 $\\$ Degree of p(x) and q(x) is the same i.e., 2. $\\$ Checking for division algorithm, $\\$ p(x) = g(x) × q(x) + r(x) $\\$ = 2( )$\\$ = $6x^2 + 2x + 2$ $\\$ Thus, the division algorithm is satisfied. $\\$ $\\$ (ii) deg q(x) = deg r(x) $\\$ Let us assume the division of $x^3 + x \ by \ \ x^2 ,$ $\\$ Here, p(x) = $x^3 + x$ $\\$ $g(x) = x^2$ $\\$ q(x) = x and r(x) = x $\\$ Clearly, the degree of q(x) and r(x) is the same i.e., 1. $\\$ Checking for division algorithm, $\\$ p(x) = g(x) $\times$ q(x) + r(x) $\\$ $x^3 + x = (x^2 ) \times x + x$ $x^3 + x = x^3 + x$ $\\$ Thus, the division algorithm is satisfied. $\\$ (iii)deg r(x) = 0 $\\$ Degree of remainder will be 0 when remainder comes to a constant. $\\$ Let us assume the division of $x^3 + 1 \ by x^2 .$ $\\$ Here, p(x) = $x 3 + 1$ $\\$ $g(x) = x^2$ $\\$ q(x) = x and r(x) = 1 $\\$ Clearly, the degree of r(x) is 0. $\\$ Checking for division algorithm $\\$ $p(x) = g(x) \times q(x) + r(x)$ $\\$ $x^3 + 1 = (x^2 ) \times x + 1$ $\\$ $x^3 + 1 = x^3 + 1$ $\\$ Thus, the division algorithm is satisfied. $\\$

9   Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:$\\$ $(i) 2x_3 + x _2 + 5x + 2; \dfrac{1}{2},1,2\\ \ \ \ \ \ 2x_3 + x _2 + 5x + 2; \dfrac{1}{2},1,-2 \\ (ii) x_3 - 4x_2 + 5x - 2; 2,1,1$

Solution :

$(i)p(x) = 2x_3 + _ 2 - 5x + 2$$\\$ Zeroes for this polynomial are $\dfrac{1}{2},1,-2$$\\$ $p(\dfrac{1}{2})=2(\dfrac{1}{2})^3+(\dfrac{1}{2})^2-5(\dfrac{1}{2})+2 =0$$\\$ $p(1)=2*1^3+1^2-5*1+2 = 0 \\ p(-2) = 2(-2)^3+(-2)^2-5(-2)+2\\ =-16+4+10+2=0$$\\$ Therefore,$\dfrac{1}{2},1,$and$-2$ are the zeroes of the given polynomial.$\\$ Comparing the given polynomial with$ ax^3 + bx^2 +cx +d$ , we obtain $a = 2$,$\\$ $b = 1, c = -5, d = 2$ We can take $\alpha =\dfrac{1}{2},\beta =1,\gamma =-2$$\\$ $ \alpha +\beta +\gamma =\dfrac{1}{2} +1+(-2)= -\dfrac{1}{2} =\dfrac{-b}{a} $$\\$ $\alpha\beta +\beta\gamma +\alpha\gamma =-\dfrac{1}{2}*1+1(-2)=\dfrac{-5}{2} =\dfrac{c}{a}\\ \alpha\beta\gamma =\dfrac{1}{2}*1*(-2)= \dfrac{-1}{1}=\dfrac{-(2)}{2} =\dfrac{-d}{a} $$\\$ Therefore, the relationship between the zeroes and the coefficients is verified. $\\$ $(ii) p (x )=x^3- 4x^2 + 5x - 2$$\\$ Zeroes for this polynomial are $2, 1, 1$$\\$ $p (2 ) = 2^3 - 4 (2 )^2 + 5 ( 2 )- 2\\ = 8-16+10-2=0\\ p(1)=1^3-4(1^2)+5(1)-2\\ =1-4+5-2=0$$\\$ Therefore, $2, 1, 1$ are the zeroes of the given polynomial.$\\$ Comparing the given polynomial with $ax^3+bx^2+cx+d$ we obtain $a=1,b=-4,c=5,d=-2$.$\\$ Verification of the relationship between zeroes and coefficient of the given polynomial$\\$ Sum of zeroes =$ 2+1+1=4=\dfrac{-(-4)}{1} =\dfrac{-b}{a}$$\\$ Multiplication of zeroes taking two at a time=$(2)(1)+(1)(1)+(2)(1)$$\\$ $=2+1+2=5=\dfrac{(5)}{1} =\dfrac{c}{a}$$\\$ Multiplication of zeroes = $2*1*1=2=\dfrac{-(-2)}{1}=\dfrac{-d}{a}$$\\$ Hence, the relationship between the zeroes and the coefficients is verified.

$(ii) p (x )=x^3- 4x^2 + 5x - 2$$\\$ Zeroes for this polynomial are $2, 1, 1$$\\$ $p (2 ) = 2^3 - 4 (2 )^2 + 5 ( 2 )- 2\\ = 8-16+10-2=0\\ p(1)=1^3-4(1^2)+5(1)-2\\ =1-4+5-2=0$$\\$ Therefore, $2, 1, 1$ are the zeroes of the given polynomial.$\\$ Comparing the given polynomial with $ax^3+bx^2+cx+d$ we obtain $a=1,b=-4,c=5,d=-2$.$\\$ Verification of the relationship between zeroes and coefficient of the given polynomial$\\$ Sum of zeroes =$ 2+1+1=4=\dfrac{-(-4)}{1} =\dfrac{-b}{a}$$\\$ Multiplication of zeroes taking two at a time=$(2)(1)+(1)(1)+(2)(1)$$\\$ $=2+1+2=5=\dfrac{(5)}{1} =\dfrac{c}{a}$$\\$ Multiplication of zeroes = $2*1*1=2=\dfrac{-(-2)}{1}=\dfrac{-d}{a}$$\\$ Hence, the relationship between the zeroes and the coefficients is verified.

$\alpha\beta +\beta\gamma +\alpha\gamma =-\dfrac{1}{2}*1+1(-2)=\dfrac{-5}{2} =\dfrac{c}{a}\\ \alpha\beta\gamma =\dfrac{1}{2}*1*(-2)= \dfrac{-1}{1}=\dfrac{-(2)}{2} =\dfrac{-d}{a} $$\\$ Therefore, the relationship between the zeroes and the coefficients is verified.

10   Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as $2, -7, - 14$ respectively.

Solution :

Let the polynomial be$ax^3+bx^2+cx+d$and the zeroes be $\alpha, \beta, $ and $\gamma$.It is given that$\alpha +\beta +\gamma =\dfrac{2}{1} = \dfrac{-b}{a}$$\\$ $\alpha\beta +\beta \gamma +\alpha \gamma =\dfrac{-7}{1} =\dfrac{c}{a} \\ \alpha \beta \gamma =\dfrac{-14}{1} =\dfrac{-d}{a} $$\\$ If $a=1$,then $b=-2,c=-7,d=14$$\\$ Hence, the polynomial is$x^3-2x^2 -7x+14$.

11   If the zeroes of polynomial$x^3-3x^2+x+1$are $a-b,a,a+b$find $a$ and $b.$

Solution :

$ p(x) =x^3 -3x^2 + x+1 $$\\$ Zeroes are $a -b, a + a + b$$\\$ Comparing the given polynomial with$px^3+qx^2+rx+t$,we obtain$\\$ $ p=1,q=-3,r=1,t=1$$\\$ Sum of zeroes=$a-b+a+a+b$$\\$ $ \dfrac{-q}{p} =3a \\ \dfrac{-(-3)}{1}=3a \\ 3=3a\\ a=1$$\\$ The zeroes are $1 -b, 1 + b.$ $\\$ Multiplication of zeroes=$1(1-b)(1+b)$$\\$ $ \dfrac{-t}{p} =1-b^2 \\ \dfrac{-1}{1} =1-b^2 \\ -1 \\ 1+1=b^2 \\b=\pm\sqrt{2} $$\\$ Hence, $a=1$ and $b=\sqrt{2}$ or $-\sqrt{2}$.

12   It two zeroes of the polynomial $x^4 -6x^3 -138x -35 $ are $2\pm\sqrt{3}$, find other zeroes.

Solution :

Given that $2 +\sqrt{ 3}$ and $2- 3$ are zeroes of the given polynomial.$\\$ Therefore, $(x-2-3)(x-2+3) = x 2 + 4 -4x -3$$\\$ $= x^2 - 4x + 1$ is a factor of the given polynomial$\\$ For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing$x^4-6x^3-26x^2+138x-35$ by $x^2-4x+1$ $\\$ $$ \require{enclose} \begin{array}{rll} x^2-2x-35&& \\[-3pt] x^2-4x+1\enclose{longdiv}{x^4-6x^3-26x^2+138x-35}\kern-.2ex \\[-3pt] {x^4-4x^3+x^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{- \quad +\quad + \qquad \qquad \ \ \phantom{00}} && \\[-3pt] -2x^3-27x^2+138x-35\phantom{0} && \\[-3pt] {\phantom{0} -2x^3+8x^2-2x \qquad \phantom{0}} && \\[-3pt] \underline{+ \quad - \quad + \qquad \quad \phantom{00}} && \\[-3pt] -35x^2+140x-35 \phantom{0} && \\[-3pt] {\phantom{0} -35x^2+140x-35 \phantom{0}} && \\[-3pt] \underline{+ \quad - \quad + \phantom{00}} && \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ 0 \ \ } && \\[-3pt] \phantom{00} \end{array}$$ $\\$ Clearly,=$x^4 -6x^3-26x^2+138x-35 =(x^2 -4x+1)(x^2-2x-35)$$\\$ It can be observed that$(x^2-2x-35)$ is also a factor of the given polynomial.$\\$ And =$(x^2- 2x-35) =(x-7)(x+5)$$\\$ Therefore, the value of the polynomial is also zero when or $x-7=0$$\\$ Or $x+5=0$$\\$ Or$x=7 or -5$$\\$ Hence,$ 7$ and $-5$ are also zeroes of this polynomial. $$

Given that$2+\sqrt{3}$ and $2-\sqrt{3 } $ are zeroes of the given polynomial.$\\$ Therefore,$(x-2-\sqrt{3})(x-2+\sqrt{3}) =x^2 +4-4X-3 =x^2-4x+1$is a factor of the given polynomial$\\$ For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing$x^4-6x^3-26x^2+138x-35$ by $x^2-4x+1$$\\$

Clearly,=$x^4 -6x^3-26x^2+138x-35 =(x^2 -4x+1)(x^2-2x-35)$$\\$ It can be observed that$(x^2-2x-35)$ is also a factor of the given polynomial.$\\$ And =$(x^2- 2x-35) =(x-7)(x+5)$$\\$ Therefore, the value of the polynomial is also zero when or $x-7=0$$\\$ Or $x+5=0$$\\$ Or$x=7 or -5$$\\$ Hence,$ 7$ and $-5$ are also zeroes of this polynomial.

$ \polylongdiv{x^4-6x^2-26x^2+138x-35}{x^2-4x+1}$$\\$ $$ \require{enclose} \begin{array}{rll} x^2 + 2x - 35 && \\[-3pt] x^2 -4x +1\enclose{longdiv}{x^4-6x^3-26x^2 +138x-35}\kern-.2ex \\[-3pt] {x^4-4x^3+x^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{- \quad +\quad -\qquad \qquad \ \ \phantom{00}} && \\[-3pt] -2x^3-27x^2+138x-35 \phantom{0} && \\[-3pt] {\phantom{0} -2x^3+8x^2-2x\qquad \phantom{0}} && \\[-3pt] \underline{+ \quad - \quad + \qquad \quad \phantom{00}} && \\[-3pt] -35x^2+140x-35 \phantom{0} && \\[-3pt] {\phantom{0} -35x^2+140x-35\phantom{0}} && \\[-3pt] \underline{+ \quad - \quad + \phantom{00}} && \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ 0 \ \ } && \\[-3pt] \phantom{00} \end{array} $$

13   If the polynomial $x^4- 6x^3+16x^2 -25x-10$is divided by another Polynomial$x^2 - 2x +k$,the remainder comes out to be$x+a$,ind $k$ and $a$.

Solution :

By division algorithm,$\\$ Dividend = Divisor × Quotient + Remainder$\\$ Dividend - Remainder = Divisor × Quotient$\\$ $ x^4-6x^3 +16 x^2 -25x -10 -x -a =6x^3 +16x^2 - 26x+10-a $will be perfectly divisible by$x^2 -2x+k$.$\\$ Let us divide by$x^4 -6x^3 -26x-10-a$ by $x^2 - 2x+k$$\\$ $$ \require{enclose} \begin{array}{rll} x^2-4x+(8-k) && \\[-3pt] x^2-2x+k\enclose{longdiv}{x^4-6x^3+16x^2-26x+10-a}\kern-.2ex \\[-3pt] {x^4-2x^3+kx^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{- \quad + \quad -\qquad \qquad \ \ \phantom{00}} && \\[-3pt] -4x^3+(16-k)x^2-26x \phantom{0} && \\[-3pt] {\phantom{0} -4x^3+8x^2-4kx\qquad \phantom{0}} && \\[-3pt] \underline{+\quad - \quad + \qquad \quad \phantom{00}} && \\[-3pt] (8-k)x^2-(26-4k)x+10-a \phantom{0} && \\[-3pt] {\phantom{0} (8-k)x^2-(16-2k)x+(8k-k^2) \phantom{0}} && \\[-3pt] \underline{- \quad + \quad -\phantom{00}} && \\[-3pt] (-10+2k)x+(10-a-8k+k^2)\phantom{0}&& \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ \ \ } && \\[-3pt] \phantom{00} \end{array} $$$\\$ $(x^2-4x+1)(x^2-2x-35) =(x-7)(x+5)$$\\$ It can be observed that $(-10+2k)x+(10-a-8k+k^2)$ will be $0$ .$\\$ Therefore, $(-10+2k)=0$ and $(10-a-8k+k^2)=0$ $\\$ For $(-10+2k) =0 ,\\ 2k=10$$\\$ And thus , $ k=5$$\\$ For$ (10-a-8k+k^2) =0\\ 10-a-8*5+25=0\\ -5-a=0$$\\$ Therefore,$a=-5$$\\$ Hence, $k=5$ and $a=-5$

It can be observed that $(-10+2k)x+(10-a-8k+k^2)$ will be $0$ .$\\$ Therefore, $(-10+2k)=0$ and $(10-a-8k+k^2)=0$ $\\$ For $(-10+2k) =0 ,\\ 2k=10$$\\$ And thus , $ k=5$$\\$ For$ (10-a-8k+k^2) =0\\ 10-a-8*5+25=0\\ -5-a=0$$\\$ Therefore,$a=-5$$\\$ Hence, $k=5$ and $a=-5$

$$ \require{enclose} \begin{array}{rll} x^2-4x+(8-k) && \\[-3pt] x^2-2x+k\enclose{longdiv}{x^4-6x^3+16x^2-26x+10-a}\kern-.2ex \\[-3pt] {x^4-2x^3+kx^2 \qquad \quad \ \phantom{00}} && \\[-3pt] \underline{- \quad + \quad -\qquad \qquad \ \ \phantom{00}} && \\[-3pt] -4x^3+(16-k)x^2-26x \phantom{0} && \\[-3pt] {\phantom{0} -4x^3+8x^2-4kx\qquad \phantom{0}} && \\[-3pt] \underline{+\quad - \quad + \qquad \quad \phantom{00}} && \\[-3pt] (8-k)x^2-(26-4k)x+10-a \phantom{0} && \\[-3pt] {\phantom{0} (8-k)x^2-(16-2k)x+(8k-k^2) \phantom{0}} && \\[-3pt] \underline{- \quad + \quad -\phantom{00}} && \\[-3pt] (-10+2k)x+(10-a-8k+k^2)\phantom{0}&& \\[-3pt] \underline{\phantom{0} \qquad \qquad \ \ \ \ \ } && \\[-3pt] \phantom{00} \end{array} $$$\\$ $(x^2-4x+1)(x^2-2x-35) =(x-7)(x+5)$