Question

Find the zeroes of the following cubic polynomial and verify the relations between zeroes and co-efficient: p(x) = 2x ^ 3 + 9x² +13x + 6​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: f(x) = {2x}^{3} + {9x}^{2} + 13x + 6$

Now, we have to find zeroes of f(x).

$\red{\rm :\longmapsto\:Let \: x = - 1}$

$\rm :\longmapsto\:f( - 1) = - 2 + 9 - 13 + 6$

$\rm :\longmapsto\:f( - 1) = - 15 + 15$

$\rm :\longmapsto\:f( - 1) = 0$

$\bf\implies \:x + 1 \: is \: a \: factor \: of \: f(x).$

So, in order to find the remaining factors, we use long division.

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: 2{x}^{2} + 7x + 6\:\:}}}\\ {\underline{\sf{x + 1}}}& {\sf{\: {2x}^{3} + {9x}^{2} + 13x + 6 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: - 2{x}^{3} - 2 {x}^{2} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 7{x}^{2} + 13x + 6 \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: - 7{x}^{2} - 7x \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: 6x + 6 \:\:}} \\{\sf{}}& \underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: - 6x -6\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \: \: \: \: \: \: \: \: \: 0\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

We know,

$\rm :\longmapsto\:Dividend = Divisor \times Quotient + Remainder$

$\rm :\longmapsto\:f(x) = (x + 1)( {2x}^{2} + 7x + 6)$

$\rm :\longmapsto\:f(x) = (x + 1)( {2x}^{2} + 4x + 3x+ 6)$

$\rm :\longmapsto\:f(x) = (x + 1)(2x(x + 2) + 3(x + 2))$

$\rm :\longmapsto\:f(x) = (x + 1)(x + 2)(2x + 1)$

$\bf\implies \:zeroes \: of \: f(x) = - 1, \: - 2, \: - \dfrac{3}{2}$

$\rm :\longmapsto\:Let \: \alpha = - 1, \: \beta = - 2, \: \gamma = - \dfrac{1}{2}$

Now,

Consider,

$\red{\rm :\longmapsto\:S_1 = Sum \: of \: zeroes \: taken \: one \: at \: time}$

$\: \: \rm \: = \: \: \alpha + \beta + \gamma$

$\: \: \rm \: = \: \: - 1 - 2 - \dfrac{3}{2}$

$\: \: \rm \: = \: \: - 3- \dfrac{3}{2}$

$\: \: \rm \: = \: \: \dfrac{ - 6 - 3}{2}$

$\: \: \rm \: = \: \: \dfrac{ - 9}{2}$

$\: \: \rm \: = \: \: - \dfrac{coefficient \: of \: {x}^{2} }{coefficient \: of \: {x}^{3} }$

$\: \: \: \: \: \red{\bf\implies \: \alpha + \beta + \gamma = - \dfrac{coefficient \: of \: {x}^{2} }{coefficient \: of \: {x}^{3} } }$

Consider,

$\blue{\bf :\longmapsto\:S_2 = Sum \: of \: two \: zeroes \: taken \: at \: time}$

$\: \: \rm \: = \: \: \alpha \beta + \beta \gamma + \gamma \alpha$

$\: \: \rm \: = \: \:( - 1)( - 2) + ( - 2)\bigg(\dfrac{ - 3}{2} \bigg) + \bigg(\dfrac{ - 3}{2} \bigg)( - 1)$

$\: \: \rm \: = \: \:2 + 3 + \dfrac{3}{2}$

$\: \: \rm \: = \: \:5 + \dfrac{3}{2}$

$\: \: \rm \: = \: \:\dfrac{10 + 3}{2}$

$\: \: \rm \: = \: \:\dfrac{13}{2}$

$\: \: \rm \: = \: \:\dfrac{coefficient \: of \: {x} }{coefficient \: of \: {x}^{3} }$

$\: \: \: \: \: \blue{\bf\implies \: \alpha \beta + \beta \gamma + \gamma \alpha = - \dfrac{coefficient \: of \: {x}}{coefficient \: of \: {x}^{3} } }$

Consider,

$\green{\bf :\longmapsto\:S_3 = Product \: of \: zeroes \: }$

$\: \: \rm \: = \: \: \alpha \beta \gamma$

$\: \: \rm \: = \: \:( - 1)( - 2)\bigg(\dfrac{ - 1}{2} \bigg)$

$\: \: \rm \: = \: \: - 3$

$\: \: \rm \: = \: \: - \dfrac{6}{2}$

$\: \: \rm \: = \: - \:\dfrac{constant \: term}{coefficient \: of \: {x}^{3} }$

$\: \: \: \: \: \green{\bf\implies \: \alpha\beta\gamma = - \dfrac{constant \: term}{coefficient \: of \: {x}^{3} } }$

Hence, Verified