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Descartes' Rule

If 1, 2, 3 ...

Question

If $1, 2, 3$ are the roots of $x^{3} + a x^{2} + b x + c = 0$ , then $(a, b, c) =$

(- 6, 11, - 6)

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(6, 11, 6)

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(- 6, 11, 6)

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(6, 11, - 6)

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Solution

The correct option is A $(- 6, 11, - 6)$
The equation with these three roots will be
$(x - 1) (x - 2) (x - 3) = 0.$
Expanding and rearranging, we get
$(x^{2} - 3 x + 2) (x - 3) = 0$
$x^{3} - 6 x^{2} + 11 x - 6 = 0$
Comparing it with $x^{3} + a x^{2} + b x + c = 0$ , we get
$a = - 6$ , $b = 11$ and $c = - 6$

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