If alpha & beta r the …

The given quadratic polynomial is 
f(x) = x² - 3x - 2
Sum of the zeroes = α + β = 3
Product of the zeroes = αβ = -2
From the question, we have the quadratic polynomial ax² + bx + c whose zeroes are {tex}\frac{1}{{2\alpha + \beta }} \space and\ \space \frac{1}{{2\beta + \alpha }}{/tex}.
Sum of the zeroes of the new polynomial {tex}= \frac{1}{{2\alpha + \beta }} + \frac{1}{{2\beta + \alpha }}{/tex}
{tex} = \frac{{2\beta + \alpha + 2\alpha + \beta }}{{\left( {2\alpha + \beta } \right)\left( {2\beta + \alpha } \right)}}{/tex}
{tex}= \frac{{3\alpha + 3\beta }}{{2\left( {{\alpha ^2} + \beta^2} \right) + 5\alpha \beta }}{/tex}
{tex} = \frac{{3 \times 3}}{{2\left[ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta + 5 \times \left( { - 2} \right)} \right]}}{/tex}
{tex}= \frac{9}{{2\left[ {9 - ( - 4)} \right] - 10}}{/tex}
{tex}= \frac{9}{{2\left[ {13} \right] - 10}}{/tex}
{tex} = \frac{9}{{26 - 10}} = \frac{9}{{16}}{/tex}
Also Product of the zeroes {tex}\frac{1}{{2\alpha + \beta }} \space \times\ \space \frac{1}{{2\beta + \alpha }}{/tex}
{tex}= \frac{1}{{\left( {2\alpha + \beta } \right)\left( {2\beta + \alpha } \right)}}{/tex}
{tex}= \frac{1}{{4\alpha \beta + 2{\alpha ^2} + 2{\beta ^2} + \alpha \beta }}{/tex}
{tex} = \frac{1}{{5\alpha \beta + 2\left( {{\alpha ^2} + {\beta ^2}} \right)}}{/tex}
{tex} = \frac{1}{{5\alpha \beta + 2\left( {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right)}}{/tex}
{tex} = \frac{1}{{5 \times \left( { - 2} \right) + 2\left( {{{\left( 3 \right)}^2} - 2 \times \left( { - 2} \right)} \right)}}{/tex}
{tex}= \frac{1}{{ - 10 + 26}} = \frac{1}{{16}}{/tex}
So, the quadratic polynomial is,
x² - (sum of the zeroes)x + (product of the zeroes)
{tex}= \left( {{x^2} + \frac{9}{{16}}x + \frac{1}{{16}}} \right){/tex}
Hence, the required quadratic polynomial is {tex}\left( x ^ { 2 } + \frac { 9 } { 16 } x + \frac { 1 } { 16 } \right){/tex}.