$alpha$ and $beta$ are the zeros of the polynomial $x^2+4x+3$. Then write the polynomial whose zeros are $1+frac{alpha}{beta}$ and $1+frac{beta}{alpha}$.Ã¢â‚¬â€¹

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 $1+\frac{\beta}{\alpha}$.

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Solution:

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

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

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

$â€‹=\frac{\alpha\beta+\beta^2 +\alpha\beta+\alpha^2}{\alpha\beta}$

$â€‹=\frac{\alpha^2 +\beta^2 +2\alpha\beta}{\alpha\beta}$

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

$= \frac{(âˆ’4)^2}{3}$

$â€‹=\frac{16}{3}$
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Also, Product of the zeros of new polynomial is

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

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

$â€‹=\frac{(\alpha+\beta)^2}{\alpha\beta}$

$=\frac{(âˆ’4)^2}{3}$

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

$\Rightarrow k\times[x^2âˆ’ 31x+ 316]$

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

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