Question

Obtain all zeroes of 3x^4+6x^3-2x^2-10x-5, if two of its zeroes are root 5/3 and -root 5/3

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Answer 1

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

Let assume that

$\rm :\longmapsto\:p(x) = {3x}^{4} + {6x}^{3} - {2x}^{2} - 10x - 5$

Given that

$\rm :\longmapsto\: - \sqrt{\dfrac{5}{3} } \: and \: \sqrt{\dfrac{5}{3} } \: are \: zeroes \: of \: p(x)$

$\rm :\longmapsto\:x - \sqrt{\dfrac{5}{3} } \: and \: x + \sqrt{\dfrac{5}{3} } \: are \: factors \: of \: p(x)$

$\rm :\longmapsto\: \bigg(x - \sqrt \dfrac{5}{3} \bigg) \bigg(x + \sqrt \dfrac{5}{3} \bigg) \: is \: factor \: of \: p(x)$

$\rm :\longmapsto\: \bigg( {x}^{2} - \dfrac{5}{3} \bigg) \: is \: factor \: of \: p(x)$

$\rm :\longmapsto\: \bigg( \dfrac{3 {x}^{2} - 5}{3} \bigg) \: is \: factor \: of \: p(x)$

$\rm :\longmapsto\:3 {x}^{2} - 5 \: is \: factor \: of \: p(x)$

Now,

Using Long Division, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: \: \: \: {x}^{2} + 2x + 1 \: \: \:}}}\\ {{\sf{ {3x}^{2} - 5}}}& {\sf{\: {3x}^{4} + {6x}^{3} - {2x}^{2} - 10x - 5 \:}} \\{\sf{}}&\underline{\sf{\: \: \: { - 3x}^{4} + \: \: \: \: \: \: \: \: \: 5{x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: {6x}^{3} + {3x}^{2} - 10x - 5 \:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: - {6x}^{3} \: \: \: \: \: \: \: \: \: + 10x \: \: \: \: \: \: \: \: \: \:\:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {3x}^{2} - 5\:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: { - 3x}^{2} + 5 \:\:}}\\{\sf{}}&\underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}$

Now, we know that,

By Euclid Division Algorithm,

Dividend = Divisor × Quotient + Remainder

So,

$\rm :\longmapsto\:{3x}^{4} + {6x}^{3} - {2x}^{2} - 10x - 5$

$\rm \: = \: \:( {3x}^{2} - 5)( {x}^{2} + 2x + 1)$

$\rm \: = \: \:( {3x}^{2} - 5)( {x}^{2} + x + x + 1)$

$\rm \: = \: \:( {3x}^{2} - 5) \bigg(x(x + 1) +1( x + 1) \bigg)$

$\rm \: = \: \:( {3x}^{2} - 5) (x + 1)(x + 1)$

Hence,

The zeroes of

$\rm :\longmapsto\:p(x) = {3x}^{4} + {6x}^{3} - {2x}^{2} - 10x - 5$

is

$\red{ \boxed{ \rm{ \: - \sqrt{\dfrac{5}{3} }, \: \: \sqrt{\dfrac{5}{3} }, \: \: - 1, \: \: - 1}}}$