Find all the zeroes of 2x-3x-3x+6x-2, if you know that two of its zeroes are √2 and -√2
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Answer:
roots of f(x) = ±√2, 1, -3/2
Step-by-step explanation:
f(x) = 2x^4-3x³-3x²+6x-2 = 0
(x-1) is a factor of f(x)
f(x) = 2x^4-3x³-3x²+6x-2
= (2x³-x²-4x+2)(x-1)
2x³-x²-4x+2 = 0
√2 and -√2 are also roots of f(x),
α+β = 0. αβ = -2
x²-(α+β)x+αβ = 0
x²-0-2 = 0
x²-2 = 0
x²-2 is a factor of f(x),
2x³-x²-4x+2 ÷ x²-2 = 2x+3
2x+3 = 0 (2x+3 is a factor)
x = -3/2
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