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Given that \sqrt{2} is a zero of the cubic polynomial 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} , find its other two zeroes.

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Answer.   \left [\frac{-2\sqrt{2}}{3}, \frac{-1}{\sqrt{2}} \right ]

Given cubic polynomials are 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}

It \sqrt{2}  is a zero of the polynomial then (x – \sqrt{2} ) is a factor of 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}

6x^3 + \sqrt{2} x^2 - 10x -4\sqrt{2} = (6x^2 + 7\sqrt{2} x + 4) (x -\sqrt{2} )

= (6x^2 + 4\sqrt{2} x + 3\sqrt{2} x + 4\sqrt{2}) (x -\sqrt{2} )

= (2x(3x + 2\sqrt{2} ) +\sqrt{2} (3x + 2\sqrt{2} )) (x -\sqrt{2} )

= (3x + 2 \sqrt{2}) (2x +\sqrt{2} ) (x -\sqrt{2} )=0

3x + 2\sqrt{2} = 0                                 2x + \sqrt{2} = 0

x=\frac{-2\sqrt{2}}{3}                                     x=\frac{-1}{\sqrt{2}}                    

Hence, \left [\frac{-2\sqrt{2}}{3}, \frac{-1}{\sqrt{2}} \right ]  are the other two zeroes.

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