Find the zeroes of the polynomial: $6 x^{3} - 5 x^{2} - 3 x + 2 = 0$

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Solution

Sum of the co efficients of $x^{3}, x^{2}, x$ and the constant = 6-5-3+2 = 0

So for x = 1 is one of the roots of the expression.

(x-1) is one factor. To find the remaining, divide $6 x^{3} - 5 x^{2} - 13 x + 2$ by (x-1)
we get,

$6 x^{3} - 5 x^{2} - 3 x + 2$
$= (x - 1) (6 x^{2} + x - 2)$
$= (x - 1) (6 x^{2} + 4 x - 3 x - 2)$
$= (x - 1) [2 x (3 x + 2) - 1 (3 x + 2)]$
$= (x - 1) (2 x - 1) (3 x + 2)$
so zeros of $6 x^{3} - 5 x^{2} - 3 x + 2$ are $1, \frac{1}{2}, \frac{- 2}{3}$

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