Transforming Roots of Polynomials

Transforming the roots of a polynomial is a technique for constructing a polynomial whose roots are related to (or transformed from) the roots of another polynomial. For example, if $f(x) = x^2 -4,$ then one polynomial whose roots are the reciprocals of the roots of $f(x)$ is $g(x) = 4x^2 - 1.$ Similarly, a polynomial whose roots are one more than the roots of $f(x)$ is $g(x) = x^2-2x -3.$

The key idea is that the roots do not have to be known in order to transform the roots, often by using the results of Vieta's formula. These techniques are most often seen in math contest problems.

The general question considered in this wiki is the following:

Let $P(x)$ be a monic polynomial with roots $r_1,r_2,\ldots,r_n.$ Suppose $s_1,s_2,\ldots,s_n$ are obtained from $r_n$ by some (simple) function, e.g. $s_i = 2r_i+3,$ or $s_i = r_i^{-2}.$ What is the monic polynomial $Q(x)$ whose roots are the $s_i$?

In general, it is understood that the roots are counted with multiplicity, and that the respective multiplicities of each root and degrees of $P(x)$ and $Q(x)$ will be equal.

Here are two easy examples.

Translation.$\hspace{0.1cm}$ Let $P(x)$ be a monic polynomial with roots $r_1,\ldots,r_n.$ What is the monic polynomial $Q(x)$ whose roots are $r_1+2,r_2+2,\ldots,r_n+2$?

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Clearly $Q(x) = P(x-2)$ will do the trick: $Q(x) = 0$ if and only if $P(x-2) = 0,$ which happens if and only if $x-2 = r_i,$ or $x=r_i+2.$ This is an informal argument, since it does not formally address repeated roots, so here is a more rigorous proof:

Write $P(x) = (x-r_1)(x-r_2)(\cdots)(x-r_n),$ and $Q(x) = \big(x-(r_1+2)\big)\big(x-(r_2+2)\big)(\cdots)\big(x-(r_n+2)\big).$ Then $P(x-2) = Q(x)$ by simple substitution $($put $x-2$ in for $x$ in the expression for $P(x))$. $_\square$

For instance, let $P(x) = x^2-4$, then $Q(x) = P(x-2) = (x-2)^2-4 = x^2-4x.$ The roots of $P(x)$ are $-2,2$ and the roots of $Q(x)$ are $0,4$, as desired.

Note that this works even for complex roots. If $P(x) = x^2+4,$ then $Q(x) = P(x-2) = (x-2)^2+4 = x^2-4x+8.$ The roots of $P$ are $\pm 2i,$ and the roots of $Q(x)$ are given by the quadratic formula: $\frac{4 \pm \sqrt{4^2-4\cdot 8}}2 = \frac{4 \pm \sqrt{-16}}2 = 2 \pm \sqrt{-4} = 2 \pm 2i,$ as desired.

Scaling.$\hspace{0.1cm}$ Let $P(x)$ be a monic polynomial with roots $r_1,\ldots, r_n.$ What is the monic polynomial $Q(x)$ whose roots are $2r_1,\ldots,2r_n?$

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This is similar: $Q(x) = 2^n P\big(\frac x2\big)$ will work.

If $P(x) = (x-r_1)(\cdots)(x-r_n),$ then $2^n P\big(\frac x2\big) = 2^n \big(\frac x2 - r_1\big)(\cdots)\big(\frac x2-r_n\big) = (x-2r_1)(\cdots)(x-2r_n),$ which is $Q(x).$ $($The last step eliminates the $2^n$ by multiplying each of the $n$ factors by $2.)$ $_\square$

For instance, let $P(x) = x^2-4.$ Then $Q(x) = 2^2 P\big(\frac x2\big) = 4\Big(\big(\frac x2\big)^2-4\Big) = x^2-16.$ The roots of $P$ are $\pm 2$ and the roots of $Q$ are $\pm 4,$ as desired. The formula works just as well if the roots of $P$ are complex.

Inversion.$\hspace{0.1cm}$ Let $P(x)$ be a monic polynomial with roots $r_1,\ldots, r_n.$ Suppose $r_i \ne 0$ for all $i.$ What is the monic polynomial $Q(x)$ whose roots are $\frac 1{r_1}, \ldots, \frac 1{r_n}?$

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Here the natural choice is $Q(x) = P\big(\frac 1x\big),$ but this is not a polynomial: if $P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0,$ $P\big(\frac 1x\big) = \frac{1}{x^n} + \frac{a_{n-1}}{x^{n-1}} + \cdots + a_0.$ However, it becomes a polynomial after multiplying by $x^n.$ So $x^nP\big(\frac 1x\big) = a_0x^n + a_1 x^{n-1} + \cdots + a_{n-1}x + 1.$ Note that this polynomial is precisely $P(x)$ with coefficients reversed. Note that $a_0 \ne 0$ because it is $(-1)^n$ times the product of the $r_i.$ So the monic polynomial we want is $Q(x) = \frac1{a_0} x^n P\left(\frac 1x\right).\ _\square$

For example, if $P(x) = x^2-4,$ the polynomial with reversed coefficients is $-4x^2+1,$ and $Q(x) = -\frac14 \big(-4x^2+1\big) = x^2-\frac 14.$ Its roots are $\pm \frac 12,$ which are the reciprocals of the roots $\pm 2$ of $P(x).$

Slightly more difficult is the problem of finding polynomials whose roots are squares of the roots of the original polynomial.

Squaring.$\hspace{0.1cm}$ Let $P(x)$ be a monic polynomial with roots $r_1,\ldots,r_n.$ What is the monic polynomial whose roots are $r_1^2,\ldots,r_n^2?$

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First note that $P(x)$ has roots $r_i$ and $(-1)^n P(-x)$ has roots $-r_i.$ Then $(-1)^n P(x)P(-x)$ is a monic polynomial of degree $2n$ whose roots are $\pm r_1, \pm r_2, \ldots, \pm r_n.$ But note that $(-1)^n P(x)P(-x)$ is an even polynomial, so it is a polynomial in $x^2$: $(-1)^nP(x)P(-x) = Q(x^2)$ for some monic $Q$ of degree $n.$ Then $Q(x)$ will be the polynomial that the problem asks for.

$($An explicit formula for $Q(x)$ is $Q(x) = (-1)^n P(\sqrt{x})P(-\sqrt{x}).$ This formula doesn't make it clear why $Q$ is a polynomial, but the previous paragraph ensures that it does.$)$

Here is an explicit example: if $P(x) = x^3-6x^2+11x-6,$ then $\begin{aligned} -P(x)P(-x) &= \big(x^3-6x^2+11x-6\big)\big(x^3+6x^2+11x+6\big) \\ &= \big(x^3+11x\big)^2 - \big(6x^2+6\big)^2 \\ &= x^6 + 22x^4 + 121x^2 - 36x^4 - 72x^2 - 36 \\ &= x^6 - 14x^4+49x^2-36, \end{aligned}$ so $Q(x) = x^3-14x^2+49x-36$ will be the polynomial with roots $r_i^2.$ As before, the roots of $P(x)$ are $1,2,3,$ and it is easy to check that the roots of $Q(x)$ are $1,4,9$ as desired. $_\square$

These techniques can be combined to produce polynomials with roots which are interesting functions of the original roots.

Let $P(x) = x^3-6x^2+11x-6.$ Call its roots $r_1,r_2,r_3.$

What is the monic cubic polynomial whose roots are $\dfrac3{r_1+1}, \dfrac3{r_2+1}, \dfrac3{r_3+1}?$

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The monic polynomial whose roots are $r_i+1$ is $\begin{aligned} P_1(x) &= P(x-1) \\ &= (x-1)^3-6(x-1)^2+11(x-1)-6 \\ &= x^3-9x^2 + 26x - 24. \end{aligned}$ The monic polynomial whose roots are $\frac1{r_i+1}$ is $P_2(x) = -\dfrac1{24} x^3P_1\left(\frac1x\right) = -\dfrac1{24}\big(-24x^3+26x^2-9x+1\big).$ The monic polynomial whose roots are $\frac3{r_i+1}$ is $P_3(x) = 27P_2\left(\frac x3\right) = -\dfrac{27}{24}\left(\dfrac{24}{27} x^3 + \dfrac{26}9 x^2 - 3x + 1\right) = x^3 - \dfrac{13}4 x^2 + \dfrac{27}8 - \dfrac98.\ _\square$

$($In fact, the roots of $P(x)$ are $1,2,3,$ so $P_3(x)$ should have roots $\frac32, 1, \frac34.$ This is easy to check directly.$)$

More generally, suppose $f(x) = \frac{g(x)}{h(x)}$ is a rational function, with $g$ and $h$ polynomials. Given a polynomial $P$ with roots $r_i,$ the way to construct the polynomial $Q$ with roots $f(r_i)$ is as follows:

The roots of $Q$ are $y_i = \dfrac{g(r_i)}{h(r_i)},$ so they are the solutions $y_i$ to the two-variable system of equations $\begin{aligned} y h(x) - g(x) &= 0 \\ P(x) &= 0. \end{aligned}$ The monic polynomial with these roots is (up to a scaling factor) the resultant $\text{Res}_x\big(yh(x)-g(x),P(x)\big).$

Let $P(x) = x^3-x+1.$ Suppose its roots are $r_1,r_2,r_3.$ Find the polynomial whose roots are $\dfrac{r_i}{r_i^2+1}.$

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We want the resultant $\text{Res}_x\big(y(x^2+1)-x,x^3-x+1\big),$ which is $\det \begin{pmatrix} y&-1&y&0&0\\0&y&-1&y&0\\0&0&y&-1&y\\1&0&-1&1&0\\0&1&0&-1&1 \end{pmatrix} = 5y^3 - 4y^2 - y + 1.$ So $Q(x) = x^3-\frac45 x^2 - \frac15 x + \frac15.$

Another way to phrase this result is the following:
Let $\alpha$ be a root of $x^3-x+1$ and let ${\mathbb Q}(\alpha)$ be the field of polynomials in $\alpha$ with rational coefficients. Then the minimal polynomial of $\dfrac{\alpha}{\alpha^2+1}$ is $Q(x) = x^3-\frac45 x^2 - \frac15 x + \frac15.$ $_\square$

The resultant is not easy to compute by hand, but is well-suited to computer computations. This is how computer algebra packages solve the problem of finding polynomials whose roots are rational functions of the roots of existing polynomials.