Transforming Roots of Polynomial

Transforming the roots of a polynomial is a technique for constructing a polynomial whose roots are transformed from of the roots of another polynomial. For example, if \(f(x) = x^2 -4,\) then the polynomial whose roots are the reciprocals of the roots of \(f(x)\) is \(g(x) = 4x^2 - 1.\) Similarly, the 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.

Let \(P(x) = a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} + \dots + a_3x^3+a_2x^2+a_1x+a_0\).
Also, suppose \(x_1,x_2,x_3,x_4,\dots , x_{n-1} , x_n\) are complex (not necessarily distinct) roots of \(P(x)=0\).

Aim: To construct a polynomial \(Q(x)\) whose roots are \(mx_1+k, mx_2+k, mx_3+k, \ldots, mx_{n-1}+k, mx_n+k\) for some constants \(m, k\).

We let \(y=mx_i+k \Rightarrow x_i=\dfrac{y-k}{m}\) and now since \(x_i\) is root of \(P(x)=0\), \(\dfrac{y-k}{m}\) is also root of \(P(x)\) and our required polynomial is

\[Q(x) = P\left(\dfrac{y-k}{m}\right) = a_n\left(\dfrac{y-k}{m}\right)^n+a_{n-1}\left(\dfrac{y-k}{m}\right)^{n-1} + \cdots + a_2\left(\dfrac{y-k}{m}\right)^2+a_1\left(\dfrac{y-k}{m}\right)+a_0.\]

Let us try to simply this a bit:

\[a_n\left(\dfrac{y-k}{m}\right)^n+a_{n-1}\left(\dfrac{y-k}{m}\right)^{n-1} + \cdots + a_2\left(\dfrac{y-k}{m}\right)^2+a_1\left(\dfrac{y-k}{m}\right)+a_0=0.\]

Multiplying both sides by \(m^n\), we have:

\[a_n(y-k)^n+a_{n-1}m(y-k)^{n-1}+\cdots + a_2m^{n-2}(y-k)^2+a_1m^{n-1}(y-k)+a_0m^n=0.\]

Thus our polynomial in cleaner form is

\[Q(x)=a_n(y-k)^n+a_{n-1}m(y-k)^{n-1}+\cdots + a_2m^{n-2}(y-k)^2+a_1m^{n-1}(y-k)+a_0m^n=0.\]

Aim: To construct a polynomial \(Q(x)\) whose roots are \(\dfrac{k}{x_1}, \dfrac{k}{x_2}, \dfrac{k}{x_3}, \dots, \dfrac{k}{x_{n-1}}, \dfrac{k}{x_n}\) for some constant \(k\) .

We let \(y=\dfrac{k}{x_i} \Rightarrow x_i=\dfrac{k}{y}\) now since \(x_i\) is root of \(P(x)=0\), \(\dfrac{k}{y}\) is also root of \(P(x)\) and our required polynomial is

\[Q(x)=P\left(\dfrac{k}{y}\right)= a_n\left(\dfrac{k}{y}\right)^n+a_{n-1}\left(\dfrac{k}{y}\right)^{n-1}+ \cdots + a_3\left(\dfrac{k}{y}\right)^3+a_2\left(\dfrac{k}{y}\right)^2+a_1\left(\dfrac{k}{y}\right)+a_0.\]

Let us try to simplify a bit:

\[a_n\left(\dfrac{k}{y}\right)^n+a_{n-1}\left(\dfrac{k}{y}\right)^{n-1}+ \cdots + a_3\left(\dfrac{k}{y}\right)^3+a_2\left(\dfrac{k}{y}\right)^2+a_1\left(\dfrac{k}{y}\right)+a_0=0.\]

Multiplying both sides by \(y^n\), we get:

\[a_0y^n+a_1ky^{n-1}+a_2k^2y^{n-2} + \cdots + a_{n-2}k^{n-2}y^2+a_{n-1}k^{n-1}y+a_nk^n=0.\]

Hence our polynomial in cleaner form is

\[Q(x)=a_0y^n+a_1ky^{n-1}+a_2k^2y^{n-2} + \cdots + a_{n-2}k^{n-2}y^2+a_{n-1}k^{n-1}y+a_nk^n=0.\]

E.g., problems like \(f(n) = n^2\) for \(n=1,2,\ldots,\) so use the fact that the roots of \(g(n) = f(n) - n^2\) are \(n=1,2,\ldots,k\) figure something out about \(f(k+1).\)