Find polynomial whose root is sum of roots of other polynomials 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?

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.
