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If $\alpha, \beta$ are the zeroes of the polynomial $3x^2 - 13x - 10$, then find the value of $(3\alpha + 1) (3\beta + 1)$.
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$\alpha, \beta$ are zeroes of $3x^2 - 13x - 10$
$\therefore \alpha + \beta = \frac{13}{3}$, $\alpha\beta = \frac{-10}{3}$
$(3\alpha + 1). (3\beta + 1) = 9\alpha\beta + 3(\alpha + \beta) + 1$
$= -30 + 13 + 1$
$= -16$
$\therefore \alpha + \beta = \frac{13}{3}$, $\alpha\beta = \frac{-10}{3}$
$(3\alpha + 1). (3\beta + 1) = 9\alpha\beta + 3(\alpha + \beta) + 1$
$= -30 + 13 + 1$
$= -16$