Question

Can someone pls help me find dis!!

Answer 1

given :-

» (2+√3) and (2-√3) are zeros of polynomial f(x) =
(2x^4-9x³+5x²+3x-1)

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So,

=> [x-(2+√3)].[x-(2-√3)] = (x-2-√3).(x-2+√3)

=> (x-2)²-√3²

=> x²-4x+4-3

=> x²-4x+1
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» we know that "x²-4x+1" is the factor of polynomial.

» then f(x) is divide by x²-4x+1

$= > \frac{2 {x}^{4} - 9 {x}^{3} + 5 {x}^{2} + 3x - 1 }{ {x}^{2} - 4x + 1 } \\ \\ = > 2 {x}^{2} - x - 1$

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So

=> f(x) = (x²-4x+1).(2x²-x-1)

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» HENCE the other two zeroes of polynomial f(x) are the zeroes of the polynomial "2x²-x-1"

=> 2x² - x - 1

=> 2x² - 2x + x - 1

=> 2x(x-1) +1(x-1)

=> (2x+1)(x-1)

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» Hence the other zeroes are "-1/2 and 1"

__________________[ANSWER]

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_-_-_-_✌☆$BE \: \: \: \: BRAINLY$☆✌_-_-_-_

Answer 2

The answer is in attachment. Your answer is -1/2 and 1.

HOPE THE ANSWER HELPS YOU .