Q.Find the equation whose roots are twice the roots of $x^3-6x^2+11x-6=0$.

Given an equation $f(x)=0$ with roots $\alpha_1,\ldots,\alpha_n$, several standard substitutions build a new equation whose roots are related to the $\alpha_i$ by a fixed rule, without ever finding the $\alpha_i$ individually.

• Change of sign: $f(-x)=0$ has roots $-\alpha_1,\ldots,-\alpha_n$.
• Multiply the roots by $k$: $f(x/k)=0$, cleared of fractions, has roots $k\alpha_1,\ldots,k\alpha_n$.
• Translate the roots (diminish/increase by $h$): the equation in $y=x-h$ is obtained by expanding $f(y+h)$. This expansion is carried out efficiently by repeated synthetic division of $f$ by $(x-h)$ -- divide, then divide the quotient by $(x-h)$ again, and again; the successive remainders, read from the last one obtained back to the first, are exactly the coefficients of the transformed equation in $y$, constant term first. This bookkeeping is Horner's process. Choosing $h=-p_1/n$ for a monic degree-$n$ equation $x^n+p_1x^{n-1}+\cdots$ removes its second-highest term entirely.
• Reciprocal-roots transformation: $x^nf(1/x)=0$, i.e. reversing the coefficients of $f$, has roots $1/\alpha_1,\ldots,1/\alpha_n$.
• Squared roots: substituting $x=\sqrt y$ into $f(x)=0$, collecting the surviving $\sqrt y$ terms on one side and squaring to eliminate the radical produces a polynomial in $y$ whose roots are $\alpha_1^2,\ldots,\alpha_n^2$.

Each transformation turns a question about the new roots into a direct coefficient computation on the old equation -- exactly what is needed both to shift or scale an equation into a more convenient form, and to solve equations whose roots satisfy an extra stated relation.