I have a polynomial $P$, with unknown roots $r_1,r_2, ... ,r_n$.
My goal is to find a polynomial $X$ with roots $s_1,s_2, ... ,s_n$, where each $s_i = 2r_i$
I shall get $X$ with no need to find the roots of $P$.
I tried to find some relation between these polynomials by using Horner's method, but I seem to be completely lost.
Thanks for help.
On
This may not be what you're looking for, but if some polynomial $p(x)$ has roots $c_1, c_2, ... , c_n$, then $q(x) = (x - 2c_1)(x - 2c_2)...(x - 2c_n)$ has root $2c_i$ for every $i \in \{1, 2, ... , n\}$.
Here's something to think about. Suppose $f(x) = x^2 + ax + b$ is a reducible polynomial. Since it is reducible, we know it has two roots, say $c_1$ and $c_2$. Since it is monic, $$f(x) = (x - c_1)(x - c_2) = x^2 + (-c_1 - c_2)x + c_1c_2$$ by the factor theorem. We can then deduce that $a = -c_1 - c_2$ and $b = c_1c_2$. If we wish to construct a polynomial with roots $2c_1$ and $2c_2$, we can construct the polynomial \begin{align*} (x - 2c_1)(x - 2c_2) &= x^2 + (-2c_1 - 2c_2)x + 2c_12c_2 \\ &=x^2 + 2(-c_1 -c_2)x + 4c_1c_2 \\ &= x^2 + 2ax + 4b \end{align*} We have now constructed a new polymial from $f(x)$ with roots $2c_1$ and $2c_2$.
If you substitute $x=ay$ (with $a$ some constant) into a polynomial $P$ in $x$, then you get a polynomial $Q$ in $y$. Moreover if $x_0$ is a root of $P$, that is setting $x=x_0$ in $P$ makes the result evaluate to $0$, then substituting $y=x_0/a$ into $Q$ will have the same effect; therefore $x_0/a$ will be a root of $Q$. Therefore what you want, doubling all the roots of $P$ is achieved by taking $a=\frac12$, so substituting $x=y/2$. Afterwards you can replace $y$ by $x$, if you want a polynomial in $x$. This operation then replaces any polynomial $$ c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0 \quad\text{by}\quad \frac{c_n}{2^n}x^n + \frac{c_{n-1}}{2^{n-1}}x^{n-1}\cdots+\frac{c_1}2x+c_0. $$ You don't need to know the roots explicitly for this.