x²-2/(2x²)²-3x³-3x²+6x-2
Answers
All the zeroes of this polynomial are :
√2, -√2, 1 and 1/2
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Given : √2 and -√2 are the zeroes of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2
To find : All the zeroes of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2
Solution : It is given that √2 and -√2 are the zeroes of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2
Therefore, (x+√2)(x-√2) i.e x² - 2 is the factor of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2.
Now, let us divide the given polynomial by x²-2 [Refer to the attachment]
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For other zeroes,
2x² - 3x + 1 = 0
By splitting middle term,
=> 2x² - 2x - x + 1 = 0
=> 2x ( x - 1 ) - 1 ( x - 1 ) = 0
=> (2x-1)(x-1) = 0
=> 2x-1 = 0 and x-1 = 0
=> x = 1/2 and 1
Therefore, all the zeroes of this polynomial, 2x⁴ -3x³ - 3x² + 6x - 2 are 1, 1/2, √2 and -√2.
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