Math, asked by Anonymous, 5 hours ago

Obtain other zeros of the polynomial
 \tt{f(x) =  {2x}^{4}  +  {3x}^{3}  - 5 {x}^{2} - 9x - 3 }
if two of it's zeros are
 \tt{ \sqrt{3}  \: and \:  -  \sqrt{3} }

Answers

Answered by mathdude500
8

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

Given that

\tt :\longmapsto\:f(x) =  {2x}^{4} +  {3x}^{3} -  {5x}^{2} - 9x - 3

and

\rm :\longmapsto\: -  \sqrt{3} \: and \:  \sqrt{3} \: are \: zeroes \: of \: f(x)

\rm :\longmapsto\:(x -  \sqrt{3} )\: and \: (x +  \sqrt{3}) \: are \: factors \: of \: f(x)

\rm :\longmapsto\:(x -  \sqrt{3} )(x +  \sqrt{3}) \: is \: factor \: of \: f(x)

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

Using Long Division Method, we have

\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: \:  \:  \:  \:  \:  \:  {2x}^{2}  \:  - \: 3x \:  + 1 \:  \:  \:  \: \:\:}}}\\ {{\sf{ {x}^{2} - 3}}}& {\sf{\: {2x}^{4} +  {3x}^{3} - {5x}^{2} -  9x  - 3  \:}} \\{\sf{}}&\underline{\sf{\: \:   \:  \: \:  { - 2x}^{4}  \:  \:  \:  \:  \:  \:  \:  \:  + 6 {x}^{2}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:}}\\{\sf{}}&{\sf{\:  \:  \:  \:  \  \:  {3x}^{3} +  {x}^{2} - 9x  - 3\:\:}}\\{\sf{}}&\underline{\sf{- {3x}^{3} \:  \:  \:  \:  \:  \:  \:  \:  + 9x \:\:}}\\{\sf{}}&{\sf{\:  \:  \:  \:  \  \: {x}^{2} - 3 \:\:}}\\{\sf{}}&\underline{\sf{\:\: \:  \: \:   - {x}^{2} + 3\:\:}}\\{\sf{}}&\underline{\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  0\:\: \:  \:  \:  \:  \:  \:  \:  \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}

So, we get

\rm :\longmapsto\:Dividend =  {2x}^{4} +  {3x}^{3} - 5 {x}^{2}  - 9x - 3

\rm :\longmapsto\:Quotient =  {2x}^{2}  - 3x + 1

\rm :\longmapsto\:Divisor =  {x}^{2}  - 3

\rm :\longmapsto\:Remainder = 0

We know that,

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

So,

\rm :\longmapsto\:{2x}^{4} +  {3x}^{3} - 5 {x}^{2}  - 9x - 3 = ( {x}^{2} - 3)( {2x}^{2} - 3x + 1)

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

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

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

Hence,

Other Zeroes of

\tt :\longmapsto\:f(x) =  {2x}^{4} +  {3x}^{3} -  {5x}^{2} - 9x - 3

\rm :\longmapsto\:are \: 1 \: and \:  \dfrac{1}{2}

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