$alpha$ and $beta$ are the zeros of the polynomial $x^2+4x+3$. Then write the polynomial whose zeros are $1+frac{alpha}{beta}$ and $1+frac{beta}{alpha}$.​



Given:$\alpha$ and $\beta$ are the zeros of the polynomial $x^2+4x+3$.

To do: To write the polynomial whose zeros are $1+\frac{\alpha}{\beta}$ and $1+\frac{\beta}{\alpha}$.

Solution:

$\because \alpha$ and $\beta$ are the zeros of the quadratic polynomial $x^2+4x+3$

Then, $\alpha+\beta=−4$, $\frac{\alpha}{\beta}=3$

Now, the of sum of the zeros of new polynomial is $=1+\frac{\alpha}{\beta}+1+\frac{\beta}{\alpha}$

$​=\frac{\alpha\beta+\beta^2 +\alpha\beta+\alpha^2}{\alpha\beta}$
 
$​=\frac{\alpha^2 +\beta^2 +2\alpha\beta}{\alpha\beta}$

$=\frac{( \alpha+\beta)^2}{\alpha\beta}$
 
$= \frac{(−4)^2}{3}$
 
$​=\frac{16}{3}$
Also, Product of the zeros of new polynomial is

$=2+ \frac{\alpha^2 +\beta^2}{\alpha\beta}$
 
$=\frac{2\alpha\beta+\alpha^2 +\beta^2}{\alpha\beta}$
 
$​=\frac{(\alpha+\beta)^2}{\alpha\beta}$

$=\frac{(−4)^2}{3}$
 
$=\frac{16}{3}$
Therefore, the required polynomial is $k\times[x^2−( sum\ of\ the\ zeros)x+product\ of\ zeros]$

$\Rightarrow k\times[x^2− 31x+ 316]$

$\Rightarrow 3\times(x^2 − 316x+ 316 )$                          $( if\ k=3)$

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