The sum of all non-integer roots of the equation $x^{5} - 6 x^{4} + 11 x^{3} - 5 x^{2} - 3 x + 2 = 0$ is:

6

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- 11

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- 5

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3

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Solution

The correct option is D $3$
We have, $x^{5} - 6 x^{4} + 11 x^{3} - 5 x^{2} - 3 x + 2 = 0$ .
$x = 1$ , is one root of equation.
Equation can be written as:
$x^{4} (x - 1) - 5 x^{3} (x - 1) + 6 x^{2} (x - 1) + x (x - 1) - 2 (x - 1) = 0 \Rightarrow (x - 1) (x^{4} - 5 x^{3} + 6 x^{2} + x - 2) = 0 \Rightarrow (x - 2) (x^{4} - 2 x^{3} - 3 x^{3} + 6 x^{2} + x - 2) = 0 \Rightarrow (x - 1) (x^{3} (x - 2) - 3 x^{2} (x - 2) + 1 (x - 2)) = 0 \Rightarrow (x - 1) (x - 2) (x^{3} - 3 x^{2} + 1) = 0$
Equation $(x^{3} - 3 x^{2} + 1 = 0)$ has all the non- integer roots.
Thus, sum of the all the non-integer roots of the above equation
$= \frac{- coeff. of x^{2}}{coeff. of x^{3}} = \frac{3}{1} = 3$

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