Question

find the all the zeors of the polynomials x³+3x²-2x-6 if two ofof its zeroes are -√2 and√2

Answer 1

Hey, dear friend ...
SOLUTION ↓
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The given Polynomial is f(x) = x^3+3x^2-2x-6 .

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 ,
$(x + \sqrt{2} )(x - \sqrt{2}) \\ \\ = > x(x - \sqrt{2} ) + \sqrt{2} (x - \sqrt{2} ) \\ = > {x}^{2} - \sqrt{2} x + \sqrt{2} x - 2 \\ = > {x}^{2} - 2 \: \: is \: \: a \: factor \: of \: given \: f(x). \\ \\ on \: \: dividing -$
f(x) = x^3 +3x^2-2x-6 by x^2 - 2 , we get ..

plzz see in attachment ..

we get ,

$f(x) = 0 \\ \\ = > ( {x}^{2} - 2)(x + 3) = 0 \\ = > (x + \sqrt{2} )(x - \sqrt{2} )(x + 3) = 0 \\ = > x + \sqrt{2} = 0 \: \: or \: \: x - \sqrt{2} = 0 \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \: \: \: x + 3 = 0 \\ \\ = > x = 0 - \sqrt{2} \: \: or \: x =0 + \sqrt{2} \: \: or \: x = 0 - 3 \\ \\ = > x = - \sqrt{2} \: \: or \: \: x = \: \: \: \: \sqrt{2} \: and \: x = - 3$
Hence , , all zeros of f(x) are -√2 , √2 or -3

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Hope it's helps you.
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