If $r$ is the root of an equation, $1/r$ will be the root of what equation?

Question

Let $f(x) = a + bx + cx^2 + dx^3$. It has a root $r$, and $a,b,c,d$ are all rational numbers.

How can we find another function that has a root of $1/r$? I feel like this is very obvious but I cannot seem to grasp anything.

Answer 1

Hint:

$$a+br+cr^2+dr^3=0\stackrel{\text{multiply by}\;r^{-3}}\implies ar^{-3}+br^{-2}+cr^{-1}+d=0\;\ldots\;\ldots$$

Answer 2

Let $h(x)$ be a function with the property that $h(1/r) = r$.

Then, if $f(r) = 0$, it follows that $f(h(1/r)) = 0$

That is, if $r$ is a root of $f$, then $1/r$ is a root of $f \circ h$.

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What is probably the intended solution can be obtained by picking a fairly simple function for $h$, computing $f(h(x))$, and inspecting the result to find a simpler formula that has $1/r$ as a root.

Answer 3

Note that we have, for $f(x)=dx^3+cx^2+bx+a$ that $$ g(x)=x^3f\left(\frac{1}{x}\right)=ax^3+bx^2+cx+d. $$

Answer 4

Following yup on lulu's comment, the polynomial $x^nf(1/x)$ is called the reciprocal polynomial of the $n$-th degree polynomial $f.$ In your case, $n=3.$ If $r \neq 0$ is a root of $f,$ then $1/r$ will be a root of the reciprocal polynomial. This polynomial comes up in the theory of err-r correcting codes, for instance. I first encountered the idea in Vera Pless' book "The Theory of Error-Correcting Codes." My copy is the second edition, and the topic came up in section 3 of chapter 4, on finite fields.

Answer 5

If the inverse of $r$ exists, you can obtain a polynomial that admits $\frac 1 r$ as a root simply by "reversing" the exponentiation power of the coefficients

In your example taking $g(x)=ax^3+bx^2+cx+d$ would work: $$ g\left(\frac 1 r\right) =\frac a{r^3}+\frac {b}{r^2}+\frac{c}{r} + d =\frac 1{r^3}\left( a+br+cr^2+dr^3\right) =\frac{f(r)}{r^3}=0 $$

In general assume you have a polynomial $P(x)=\sum_{i=0}^np_ix^i$

Then you can take polynomial $Q(x)=\sum_{i=0}^np_ix^{n-i}$ and you obtain for $x\neq 0$: $$ Q\left(\frac 1 x\right) =\sum_{i=0}^np_i\left(\frac 1 x\right)^{n-i} =\sum_{i=0}^np_i x^{i-n} =\frac 1{x^n}\sum_{i=0}^np_i x^i =\frac{P(x)}{x^n} $$