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If $\alpha, \beta$ are the zeroes of the polynomial $p(x) = x^2 - 2x - 3$, then find a polynomial where zeroes are $(2\alpha + 3\beta)$ and $(3\alpha + 2\beta)$.
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$p(x) = x^2-2x-3$
Here $\alpha + \beta = 2$ and $\alpha\beta = - 3$
Let the required polynomial be $x^2 - Sx + P$
$S = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5(\alpha + \beta)$
$= 5 \times 2 = 10$
$P = (2\alpha + 3\beta) \times (3\alpha + 2\beta)$
$= 6 (\alpha^2 + \beta^2) + 13 \alpha\beta$
$= 6 [(\alpha + \beta)^2 - 2 \alpha\beta] + 13 \alpha\beta$
$= 6 (\alpha + \beta)^2 + \alpha\beta$
$= 6 (2)^2 + (-3)$
$= 21$
So, required polynomial is $x^2 - 10x + 21$
Here $\alpha + \beta = 2$ and $\alpha\beta = - 3$
Let the required polynomial be $x^2 - Sx + P$
$S = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5(\alpha + \beta)$
$= 5 \times 2 = 10$
$P = (2\alpha + 3\beta) \times (3\alpha + 2\beta)$
$= 6 (\alpha^2 + \beta^2) + 13 \alpha\beta$
$= 6 [(\alpha + \beta)^2 - 2 \alpha\beta] + 13 \alpha\beta$
$= 6 (\alpha + \beta)^2 + \alpha\beta$
$= 6 (2)^2 + (-3)$
$= 21$
So, required polynomial is $x^2 - 10x + 21$