Question

find all the zeroes of 2x^4 - x^3-3x^2+6x-2, if you know that two of its zeroes are √2 and -√2

Answer 1

${2x}^{4} - {x}^{3} - 3 {x}^{2} + 6x - 2 = 0 \\ {2x}^{4 } - {x}^{3} - 3 {x}^{2} + 6x = 2 \\ {x}^{4} - {x}^{3} - {x}^{2} + x = \frac{2}{36} = \frac{1}{18} \\ x - {x}^{2} + x = \frac{1}{18 } \\ - {x}^{2} + x = \frac{1}{18 } \\ - x \times x + x = \frac{1}{18} \\ x = \frac{1}{18}$
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Answer 2

Answer:

Step-by-step explanation: