Question

Solve it 18.
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Answer 1

Heya dear ,

here is your answer,

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18)

$({assume \: \: a = alpha \: \: .b = beta \: \: .y = gama})$

$let \: a = \sqrt{2} \: . \: b = - \sqrt{2} \: be \: the \: given \: zeros \: and \: y \: be \: the \: third \: zero \: of \: f(x) \: . \: \\ then \: \\ \\ a \: + b + y = - ( \frac{ - 1}{2} ) \\ \\ = > \sqrt{2} - \sqrt{2 } + y = \frac{1}{2} \\ \\ = > y = \frac{1}{2} \\ \\ hence \: third \: zero \: is \: \frac{1}{2}$

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Hope it helps u :))

# Nikky

Answer 2

Hello dear user .

Solution is given below.
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The given Polynomial is f(x) = 2x^3 - x^2 - 4x +2.
since √2 and -√2 are the zeros of f(x) , it follows that each one of (x - √2 ) and (x+√2) is a factor of f(x).

Consequently ,
( x - √2 )( x + √2) = (x^2 -2), is a factor of f(x).

on dividing f(x) = 2x^3 - x^2 - 4x + 2 by (x^2 - 2) , we get

plzz see in the above attachment ⬆⬆⬆
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$f(x) = 0 \\ = > ( {x}^{2} - 2)(2x - 1) = 0 \\ = > (x - \sqrt{2} )(x + \sqrt{2} )(2x - 1) = 0 \\ = > (x - \sqrt{2} ) = 0 \: \: \: \: or \: \: (x + \sqrt{2} ) = 0 \\ and \: \: (2x - 1) = 0 \\ \\ = > x = \sqrt{2} \: \: or \: \: x = - \sqrt{2} \: \: \: and \: \: x = \frac{1}{2} . \\ \\ hence \: \: all \: \: zeros \: of \: f(x) \: \: are \: \sqrt{2} \: \: or \: - \sqrt{2} \: \: nd \: \frac{1}{2}$

THEREFORE third zeros of the given Polynomial is 1/2
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Hope it's helps you.
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