If $α$ and $β$ are the zeroes of the polynomial $p (x) = 3 x^{2} + 2 x + 1,$ find the polynomial whose zeroes are $\frac{1 - α}{1 + α}$ and $\frac{1 - β}{1 + β}$

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Solution

$\frac{(1 - α) (1 - β)}{(1 + α) (1 + β)} \frac{(1 - α - β + α β)}{(1 + α + β = α β)} (\frac{1 - (α + β) + α β}{1 + (α + β) + α β}) ⎡ ⎢ ⎢ ⎢ ⎣ \frac{1 + \frac{2}{3} + \frac{1}{3}}{1 - \frac{2}{3} + \frac{1}{3}} ⎤ ⎥ ⎥ ⎥ ⎦ [\frac{2 \times 3}{2}] = 3 = \frac{(1 - α) (1 + β) + (1 - β) (1 + α)}{(1 + α + β + α β)} = \frac{1 + β - α - α β + 1 + α - β - α β}{\frac{2}{3}} = \frac{3 (2 - 2 α β)}{2} = 3 (1 - α β) = 3 (1 - \frac{1}{3}) = 2$

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