If 3+i and 2-sqrt{3} are the roots of the equ | vedclass

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(B) Given $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$ with $a_i \in \mathbb{Z}$.Since the coefficients are integers,complex roots must occur in conjugate

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$4, 10$

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(B) Given $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$ with $a_i \in \mathbb{Z}$.Since the coefficients are integers,complex roots must occur in conjugate pairs and irrational roots of the form $a+\sqrt{b}$ must occur with their conjugates $a-\sqrt{b}$.Given roots are $x_1 = 3+i$ and $x_2 = 2-\sqrt{3}$.Therefore,their conjugates $x_3 = 3-i$ and $x_4 = 2+\sqrt{3}$ must also be roots.Since there are at least $4$ roots,the minimum degree $n$ is $4$.For a polynomial of degree $4$,the constant term $a_0$ is given by the product of the roots multiplied by $(-1)^n$ (if the leading coefficient is $1$): $a_0 = x_1 x_2 x_3 x_4$.$a_0 = (3+i)(3-i) 	imes (2-\sqrt{3})(2+\sqrt{3})$$a_0 = (3^2 - i^2) 	imes (2^2 - (\sqrt{3})^2)$$a_0 = (9+1) 	imes (4-3)$$a_0 = 10 	imes 1 = 10$.Thus,$n=4$ and $a_0=10$.

If $3+i$ and $2-\sqrt{3}$ are the roots of the equation $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_{n} x^{n}$ where $a_0, a_1, \ldots, a_{n} \in \mathbb{Z}$,then the least value of $n$ and the value of $a_0$ are respectively:

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