Question

Find all zeroes of 2x^4-3x^3-5x^2+9x-3 if two of its zeroes √3 and -√3

Answer 1

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Answer 2

Hello dear friend .

your Solution Given here ⬇⬇
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Let f (x) = 2x^4 - 3x^3 - 5x^2 + 9x-3
since √3 and -√3 are the zeroes of f(x)
it follows that ( x - √3 ) and ( x + √3 ) are the factor of f(x)

Consequently = ( x - √3 ) ( x + √3 )
=> x( x + √3 ) - √3( x + √3 )
=> x^2 + √3x - √3x + 3
=> x^2 - 3

So , x^2 - 3 is a factor of f (x)
dividing f(x) by x ^2 - 3

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we get remainder 0

$= > f(x) = 0 \\ = > (2 {x}^{2} - 3x + 1)(x - \sqrt{3} )(x + \sqrt{3}) = 0 \\ = > (2 {x}^{2} - 2x - 1x + 1)(x - \sqrt{3} )(x + \sqrt{3} ) = 0 \\ = > (2x(x - 1) - 1(x - 1))(x - \sqrt{3} )(x + \sqrt{3} ) = 0$
=> 2x - 1 =0 , x - 1 =0 , x - √3 =0 or x+ √3 =0

=> x = 1/2 ,. x = 1 , x = √3 and x = -√3

Hence , the zeroes of given pol are
1/2 , 1 , √3 and -√3

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