Given that √2 is a zera of the cubic polynomia
6x^2+√2x^2-10x-4√2 find its other two zeroes
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√2 is zero of 6x³ +√2x² -10x -4√2 ,
So, (x - √2) is a factor of 6x³ + √2x² -10x -4√2 .
6x³ + √2x² -10x -4√2
= 6x³ -6√2x² + 7√2x² -14x + 4x - 4√2
= 6x²(x - √2) + 7√2x(x -2) + 4(x -√2)
= {6x² + 7√2x + 4}(x - √2)
= {6x² + 3√2x + 4√2x + 4}(x -√2)
= {3√2x(√2x + 1) + 4(√2x +1)}(x -√2)
= (3√2x +4)(√2x +1)(x - √2)
Hence, two other zeros are -4/3√2, and -1/√2
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