It is known that two roots of the non-linear equation $x^{3} - 6 x^{2} + 11 x - 6 = 0$ are $1$ and $3$ . The third root will be

j

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- j

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2

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4

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Solution

The correct option is C $2$
$x^{3} - 6 x^{2} + 11 x - 6 = 0$
$x^{3} - x^{2} - 5 x^{2} + 5 x + 6 x - 6 = 0$
$x^{2} (x - 1) - 5 x (x - 1) + 6 x - 1) = 0$
$(x - 1) - 5 x (x - 1) + 6 (x - 1) = 0$
$(x - 1) (x^{2} - 5 x + 6) = 0$
$(x -) (X - 3) (x - 2) = 0$
$x = 1, 2, 3$

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