if alpha and beeta are the …

Since {tex}\alpha \text { and } \beta{/tex} are the zeroes of polynomial 3x² + 2x + 1.
Hence, {tex}\alpha + \beta = - \frac { 2 } { 3 }{/tex}
and {tex}\alpha \beta = \frac { 1 } { 3 }{/tex}
Now, for the new polynomial,
Sum of zeroes = {tex}\frac { 1 - \alpha } { 1 + \alpha } + \frac { 1 - \beta } { 1 + \beta }{/tex}
{tex}= \frac { ( 1 - \alpha + \beta - \alpha \beta ) + ( 1 + \alpha - \beta - \alpha \beta ) } { ( 1 + \alpha ) ( 1 + \beta ) }{/tex}
{tex}= \frac { 2 - 2 \alpha \beta } { 1 + \alpha + \beta + \alpha \beta } = \frac { 2 - \frac { 2 } { 3 } } { 1 - \frac { 2 } { 3 } + \frac { 1 } { 3 } }{/tex}
{tex}\therefore{/tex} Sum of zeroes = {tex}\frac { 4 / 3 } { 2 / 3 } = 2{/tex}
Product of zeroes = {tex}\left[ \frac { 1 - \alpha } { 1 + \alpha } \right] \left[ \frac { 1 - \beta } { 1 + \beta } \right]{/tex}
{tex}= \frac { ( 1 - \alpha ) ( 1 - \beta ) } { ( 1 + \alpha ) ( 1 + \beta ) }{/tex}

{tex}=\frac{1-(\alpha+\beta)+\alpha \beta}{1+(\alpha+\beta)+\alpha \beta}{/tex}
{tex}{/tex} {tex}={/tex} {tex}\frac { 1 + \frac { 2 } { 3 } + \frac { 1 } { 3 } } { 1 - \frac { 2 } { 3 } + \frac { 1 } { 3 } } = \frac { \frac { 6 } { 3 } } { \frac { 3 } { 3 } } = 3{/tex}
Hence, Required polynomial = x² - (Sum of zeroes)x + Product of zeroes
= x² - 2x + 3