Remainder factor theorem: me reason this out

[Stephen Tashi]

Can we characterize the particular numerical values that can be roots of $f(x)$ ?

If $r_1$ is a root of $f(x)$ then the roots of $2x^3 + x = r_1$ are also roots of $f(2x^3 + x)$
If it happens that $2x^3 + x = r_1$ has 3 identical roots all equal to $r_1$ then no new roots are implied. However, if $2x^3 + x = r_1$ were to have 3 distinct roots $r_2,r_3,r_4$ then the solutions to each of the equations $2x^3 + x = r_2,\ 2x^3 +x = r_3,\ 2x^3 + x = r_4$ would also be roots of $f(2x^3 + x)$.

So some numerical values $r_1$ might lead to cascade of other roots that would exceed 3000 total roots.

 

[Terrell]

Can we characterize the particular numerical values that can be roots of $f(x)$ ?

If $r_1$ is a root of $f(x)$ then the roots of $2x^3 + x = r_1$ are also roots of $f(2x^3 + x)$
If it happens that $2x^3 + x = r_1$ has 3 identical roots all equal to $r_1$ then no new roots are implied. However, if $2x^3 + x = r_1$ were to have 3 distinct roots $r_2,r_3,r_4$ then the solutions to each of the equations $2x^3 + x = r_2,\ 2x^3 +x = r_3,\ 2x^3 + x = r_4$ would also be roots of $f(2x^3 + x)$.

So some numerical values $r_1$ might lead to cascade of other roots that would exceed 3000 total roots.

the problem only hinted it as being a monic polynomial with integer coefficients and its leading term raised to 1000.

 

[Stephen Tashi]

the problem only hinted it as being a monic polynomial with integer coefficients and its leading term raised to 1000.

But (whether it's useful or not) do you see what I'm saying? For example if $f(x)$ has the factor $(x-1)$ and root $r = 1$ then $f(2x^3 + x)$ has the factor $(2x^3 + x - 1)$ so the roots of $2x^3 + x - 1 = 0$ are roots of $f(2x^3 + x -1)$. And $r = 1$ is also a root of $f(2x^3 + x)$ since we are assuming $f(x)$ is a factor of $f(2x^3 + x)$

 

[Terrell]

But (whether it's useful or not) do you see what I'm saying? For example if $f(x)$ has the factor $(x-1)$ and root $r = 1$ then $f(2x^3 + x)$ has the factor $(2x^3 + x - 1)$ so the roots of $2x^3 + x - 1 = 0$ are roots of $f(2x^3 + x -1)$. And $r = 1$ is also a root of $f(2x^3 + x)$ since we are assuming $f(x)$ is a factor of $f(2x^3 + x)$

hmm... interesting insight. i will keep that in mind. thank you!