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We have two numbers $\alpha$ and $\beta$.
We know that $\alpha$ is root of polynomial $P_n(x)$ of degree $n$ and $\beta$ is root of polynomial $Q_m(x)$ of degree $m$.

How do you find polynomial $R_{n m}(x)$ which has root equal to $\alpha+\beta$ without finding values of roots?

All polynomials are with integer coefficients.

One more question, can it be found using matrix determinant?

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If you have two polynomials, say $P(x)$ and $Q(y)$ then, their Resultant, $\operatorname{Res}(P,Q)$ is the determinant of their Sylvester matrix. It equals to $$ \operatorname{Res}(P,Q)=\prod_{P(\alpha)=0, \ P(\beta)=0,}(\alpha-\beta), $$ the product of all the differences of their roots.

Now if you consider the two variable polynomial $Q(z-y)$ as a polynomial in the variable $y$ (and therefore its roots are $\{z-\beta: \ Q(\beta)=0\}$) then, the resultant $\operatorname{Res}(P,Q)$ is a polynomial in the variable $z$ with roots $\alpha+\beta$ where $\alpha$ runs over the roots of $P$ and $\beta$ runs over the roots of $Q$.

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Of course, $x-\alpha-\beta$ has $\alpha+\beta$ as a root, but let's say that $P$ and $Q$ have rational coefficients, and we want $R$ to as well. If the roots of $P$ are $\alpha_1,\ldots,\alpha_n$, and the roots of $Q$ are $\beta_1,\ldots,\beta_n$, then $\prod_{i,j} (x-\alpha_i - \beta_j)$ has rational coefficients (they are fixed by the Galois group) and has $\alpha+\beta$ as a root.

In practice, this approach leads to working with elementary symmetric polynomials, as in the other answer, but I find this view more intuitive.

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The elementary symmetric polynomials in the $\alpha_i+\beta_j$ are symmetric in the $\alpha_i$ hence can be expressed in the elementary symmetric polynomials in the $\alpha_i$, where the coefficients are symmetriuc in the $\beta_j$, hence can be expressed in the elementary symmetric polynomials in the $\beta_j$. For details read about resultants.

Example: If $P(x)=x^2+ax+b$ and $Q(x)=x^3+cx^2+dx+e$, then $-a=\alpha_1+\alpha_2$, $b=\alpha_1\alpha_2$, $-c=\beta_1+\beta_2+\beta_3$, $d=\beta_1\beta_2+\beta_1\beta_3+\beta_2\beta_3$, $e=-\beta_1\beta_2\beta_3$. Now the coefficients of the polynomial $R(x) = x^6+fx^5+gx^4+hx^3+ix^2+jx+k$ can be found like this: For example $-f=(\alpha_1+\beta_1)+(\alpha_1+\beta_2)+(\alpha_1+\beta_3)+(\alpha_2+\beta_1)+(\alpha_2+\beta_2)+(\alpha_2+\beta_3)=3(\alpha_1+\alpha_2)+2(\beta_1+\beta_2+\beta_3)$, hence $f=3a+2c$. And $k=(\alpha_1+\beta_1)(\alpha_1+\beta_2)(\alpha_1+\beta_3)(\alpha_2+\beta_1)(\alpha_2+\beta_2)(\alpha_2+\beta_3)$, which can be transformed to $\alpha_1^3\alpha_2^3+\alpha_1^2\alpha_2^2(\alpha_1+\alpha_2)(\beta_1+\beta_2+\beta_3)+\ldots=b^3+b^2ac+\ldots $, and the other coefficients are also symmetric, but the expressions tend to grow a bit.

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