Transformation of Roots
Let us try to understand the main concept with the following example:
Example
Given a polynomial \( p(x) = x^3  2x^2 + x  1 \). Let \( \alpha, \beta \) and \( \gamma \) be its roots. Then find the polynomial equation whose roots are \( 2\alpha  3, 2\beta  3 \) and \( 2\gamma  3 \).
Observe that each root got transformed in the same way. \(alpha\) in the given polynomial is \( 2\alpha  3\) in the e=required polynomial, \(\beta\) is \(2\beta  3 \) and \(\gamma\) is \( 2\gamma  3 \).
Here the function that transforms each root is \( f(x) = 2x  3 \) in which if we put \( \alpha, \beta \) or \( \gamma \) we get \( 2\alpha  3, 2\beta  3 \) and \( 2\gamma  3 \).
Let \( y = 2x  3 \) \( \Rightarrow y + 3 = 2x \) \( \Rightarrow x = \frac{y + 3}{2} \).
Starting with the polynomial \( p(x) = x^3  2x^2 + x  1 \), and given that its roots are \( \alpha, \beta, \gamma \), we seek a new polynomial whose roots are \( 2\alpha  3, 2\beta  3, 2\gamma  3 \).
We replace \( x \) with \( (y+3)/2 \) in \( p(x) \) to match the transformed roots.
Why would this work?
We know that \(x\) is either equal to \(\alpha\), \(\beta\) or \(\gamma\). \(x\) clearly satisfies the given equation. On the other hand, we can see that \(y\) is either equal to \(2\alpha3\), \(2\beta3\) or \(2\gamma3\). By replacing \(x\) by \(\frac{y+3}{2}\) we get another cubic equation in \(y\). Clearly this new equation will have the roots \(2\alpha3\), \(2\beta3\) and \(2\gamma3\).
Expanding \( p(x) \) with \( x \) substituted gives us a new equation in terms of \( y \):
After multiplying through by 8 to clear the fractions, the simplified polynomial in terms of \( y \) is:
Since \( y \) represents \( 2\alpha  3 \), \( 2\beta  3 \), and \( 2\gamma  3 \), this is the polynomial equation that has the desired transformed roots.
Transformation of Polynomials.
Suppose that \( \alpha_1, \alpha_2, \alpha_3, ..., \alpha_n \) are the roots of a polynomial \( p(x) \) of degree \( n \). Now we wish to find another polynomial whose roots are obtained from the roots of the given polynomial through some transformation.
Follow these steps:

Identify the transformation function \( f(x) \) that will be applied to each root of the given polynomial. The function \( f(x) \) is an expression in terms of \( x \).

Given the roots \( \alpha, \beta, \gamma \) of the original polynomial equation, the roots of the required polynomial equation will be \( f(\alpha), f(\beta), \) and \( f(\gamma) \).

Solve the equation \( y = f(x) \) to express \( x \) in terms of \( y \), resulting in a new function \( g(y) \) which is the inverse of \( f(x) \). In other words, \( x = g(y) \).

Substitute \( x \) with \( g(y) \) into the original polynomial equation. The resultant equation will be in terms of \( y \) and is the required polynomial whose roots are the transformed values \( f(\alpha), f(\beta), \) and \( f(\gamma) \).

Simplify the resulting equation to find the new polynomial in terms of \( y \). This polynomial has the roots corresponding to the transformed roots by the function \( f(x) \).
By following this process, you utilize the transformation function to find a new set of roots and, consequently, a new polynomial equation that reflects this transformation.
Example
Problem: Given the polynomial equation \( p(x) = x^3  2x^2  1 \) with roots \( \alpha, \beta, \gamma \), we aim to find a new polynomial equation whose roots are \( \frac{\alpha  1}{\alpha + 2}, \frac{\beta  1}{\beta + 2}, \frac{\gamma  1}{\gamma + 2} \).
Solution: The transformation applied to the roots of the given polynomial is represented by the function \( f(x) = \frac{x  1}{x + 2} \).
Step 1: Find the Inverse Function To determine the inverse of the transformation function \( f(x) \), we set \( y = \frac{x  1}{x + 2} \) and solve for \( x \) in terms of \( y \):
The inverse function \( g(y) \) is therefore \( g(y) = \frac{2y + 1}{1  y} \).
Step 2: Substitute and Simplify Next, we substitute \( x \) with \( g(y) \) into the original polynomial \( p(x) \), leading to:
Simplifying this expression yields the new polynomial in terms of \( y \):
Step 3: Derive the New Polynomial Multiplying through by \( (1  y)^3 \) to clear the fractions and simplify, we find the required polynomial equation in \( y \):
This polynomial has the roots \( \frac{\alpha  1}{\alpha + 2}, \frac{\beta  1}{\beta + 2}, \frac{\gamma  1}{\gamma + 2} \), as desired.
Example
Problem: Given a cubic equation \( x^3  px^2 + qx  r = 0 \) with roots \( \alpha, \beta, \gamma \), find the transformed cubic equation whose roots are \( \frac{\alpha}{\beta + \gamma}, \frac{\beta}{\gamma + \alpha} \), and \( \frac{\gamma}{\alpha + \beta} \).
Solution: Using Vieta's formulas, we know that \( \alpha + \beta + \gamma = p \). The transformed roots can be expressed as \( \frac{\alpha}{p  \alpha}, \frac{\beta}{p  \beta} \), and \( \frac{\gamma}{p  \gamma} \) respectively, by substituting \( \beta + \gamma \), \( \gamma + \alpha \), and \( \alpha + \beta \) with \( p  \alpha \), \( p  \beta \), and \( p  \gamma \).
The transformation function for the roots is then \( f(x) = \frac{x}{p  x} \).
Finding the Inverse Transformation: Let \( y = f(x) \) gives us \( y = \frac{x}{p  x} \), which simplifies to \( py  xy = x \), and further to \( x = \frac{py}{y + 1} \).
Substituting into the Original Equation: Now, replace \( x \) in the original polynomial with the inverse transformation to get the new polynomial in \( y \):
Simplification: Expanding and simplifying the above expression leads us to the transformed cubic equation in \( y \):
This is the required cubic polynomial whose roots correspond to the transformed roots \( \frac{\alpha}{\beta + \gamma}, \frac{\beta}{\gamma + \alpha} \), and \( \frac{\gamma}{\alpha + \beta} \).
Example
Problem: Given the cubic equation \( x^3  x  1 = 0 \) with roots \( \alpha, \beta, \gamma \), determine the polynomial equation with roots \( (\alpha  \beta)^2, (\beta  \gamma)^2 \), and \( (\gamma  \alpha)^2 \).
Solution: Notice that \( (\alpha  \beta)^2 = (\alpha + \beta)^2  4\alpha\beta \) can be rewritten using the sum and product of the roots from the original cubic equation. Since we have \( \alpha + \beta + \gamma = 0 \) (by Vieta's formulas for a cubic equation with leading coefficient 1 and the \( x^2 \) term missing), we can express \( (\alpha  \beta)^2 \) in terms of \( \gamma \):
Substituting the known value of \( \gamma \) from the cubic equation, we get:
Identifying the transformation function as \( f(x) = \frac{2x + 1}{x} \), we find the inverse of \( f \) is found by setting \( y = \frac{2x + 1}{x} \), leading to \( xy = 2x + 1 \), which simplifies to \( x = \frac{1}{y  2} \) after rearranging.
Plugging this into the cubic equation, we obtain:
Expanding and simplifying this equation yields:
To clear the fraction, we multiply through by \( (y  2)^3 \), resulting in:
Further expanding and simplifying, we end up with a cubic equation in terms of \( y \):
This is the required cubic polynomial whose roots are the squares of the differences of the original roots: \( (\alpha  \beta)^2, (\beta  \gamma)^2 \), and \( (\gamma  \alpha)^2 \).
Special Cases
Given a polynomial equation \( p(x) = a_0x^n + a_1x^{n1} + \ldots + a_n = 0 \) with roots \( \alpha, \beta, \gamma, \ldots \), various transformations can be applied to generate new polynomial equations with transformed roots:

Negated Roots: For the polynomial equation with roots \( \alpha, \beta, \gamma, \ldots \), substitute \( x \) for \( x \) in \( p(x) \). This yields the polynomial:
\[ p(x) = a_0(x)^n + a_1(x)^{n1} + \ldots + a_n. \]The sign of the terms will alternate based on the power of \( x \). If \( n \) is even, \( (x)^n = x^n \), and if \( n \) is odd, \( (x)^n = x^n \). This transformation effectively reflects the roots of the original polynomial across the yaxis on the complex plane.

Reciprocal Roots: To find the equation whose roots are the reciprocals \( 1/\alpha, 1/\beta, 1/\gamma, \ldots \), replace \( x \) with \( 1/x \) and then multiply through by \( x^n \) to clear the denominators, resulting in:
\[ x^np(1/x) = a_0 + a_1x + \ldots + a_nx^n = 0. \]This equation has the same roots as the original equation, except they are inverted. Geometrically, it maps each root to its reciprocal on the complex plane.

Scaled Roots: For roots scaled by a constant factor \( k \), \( k\alpha, k\beta, k\gamma, \ldots \), substitute \( x/k \) into \( p(x) \) to obtain:
\[ p(x/k) = a_0(x/k)^n + a_1(x/k)^{n1} + \ldots + a_n = 0. \]This process is known as "multiplying the roots by \( k \)". Each term of the polynomial is divided by a power of \( k \), effectively scaling the roots by \( k \).

Translated Roots: For roots translated by a constant \( h \), \( \alpha  h, \beta  h, \gamma  h, \ldots \), substitute \( x + h \) for \( x \) in \( p(x) \), resulting in:
\[ p(x + h) = a_0(x + h)^n + a_1(x + h)^{n1} + \ldots + a_n = 0. \]This transformation shifts the graph of the polynomial horizontally by \( h \) units. If \( h > 0 \), the roots are shifted to the left, and if \( h < 0 \), to the right. Each term in the expansion will involve a combination of \( x \) and \( h \), impacting all coefficients except the leading one.
Given the polynomial \( p(x) = x^3  4x^2 + 5x + 6 \), we will illustrate each of the special cases with transformations and provide the resulting polynomial for each case.
For instance,
Given \( p(x) = x^3  4x^2 + 5x + 6 \):

Negated Roots: \( p(x) = x^3  4x^2  5x + 6 = 0 \) has roots opposite in sign to those of \( p(x) \).

Reciprocal Roots: \( x^3p(1/x) = 6x^3 + 5x^2  4x + 1 = 0 \) has roots that are the reciprocals of those of \( p(x) \).

Scaled Roots (k=2): \( p(x/2) = \frac{1}{8}x^3  x^2 + \frac{5}{2}x + 6 = 0 \) has roots that are twice the magnitude of those of \( p(x) \).

Translated Roots (h=1): \( p(x  1) = x^3  7x^2 + 17x  15 = 0 \) has roots that are each increased by \( 1 \) compared to those of \( p(x) \).