Question

Find all the zeros of 2x^4-3x^3+6x-8 if you know that two of its zeros are √2 and -√2​

Answer 1

Step-by-step explanation:

(x - √2)(x + √2) = 0

x² - 2 = 0

______________

x² - 2 √ 2x⁴ - 3x³ + 6x - 8 \_ 2x² -3x + 4

-2x⁴ + 4x²

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-3x³ + 4x² + 6x - 8

+3x³ -6x

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4x² - 8

-4x² + 8

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0 0

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2x² - 3x + 4 = 0

x = 3 + √9 - 4(2)(4) / 2(2)

x = 3 + √9 - 32 / 4

x = 3 + √23 i / 4

So , all the zeroes are √2 , - √2 , (3 + √23i)/4

& (3 - √23i)4