Polynomial transformation

Translating the roots Let : P(x) = a_0x^n + a_1 x^{n-1} + \cdots + a_{n} be a polynomial, and :\alpha_1, \ldots, \alpha_n be its complex roots (not necessarily distinct). For any constant , the polynomial whose roots are :\alpha_1+c, \ldots, \alpha_n+c is :Q(y) = P(y-c)= a_0(y-c)^n + a_1 (y-c)^{n-1} + \cdots + a_{n}. If the coefficients of are integers and the constant c=\frac{p}{q} is a rational number, the coefficients of may be not integers, but the polynomial has integer coefficients and has the same roots as . A special case is when c=\frac{a_1}{na_0}. The resulting polynomial does not have any term in . Reciprocals of the roots Let : P(x) = a_0x^n + a_1 x^{n-1} + \cdots + a_{n} be a polynomial. The polynomial whose roots are the reciprocals of the roots of is its reciprocal polynomial : Q(y)= y^nP\left(\frac{1}{y}\right)= a_ny^n + a_{n-1} y^{n-1} + \cdots + a_{0}. Scaling the roots Let : P(x) = a_0x^n + a_1 x^{n-1} + \cdots + a_{n} be a polynomial, and be a non-zero constant. A polynomial whose roots are the product by of the roots of is :Q(y)=c^nP\left(\frac{y}{c} \right) = a_0y^n + a_1 cy^{n-1} + \cdots + a_{n}c^n. The factor appears here because, if and the coefficients of are integers or belong to some integral domain, the same is true for the coefficients of . In the special case where c=a_0, all coefficients of are multiple of , and \frac{Q}{c} is a monic polynomial, whose coefficients belong to any integral domain containing and the coefficients of . This polynomial transformation is often used to reduce questions on algebraic numbers to questions on algebraic integers. Combining this with a translation of the roots by \frac{a_1}{na_0}, allows to reduce any question on the roots of a polynomial, such as root-finding, to a similar question on a simpler polynomial, which is monic and does not have a term of degree . For examples of this, see Cubic function § Reduction to a depressed cubic or Quartic function § Converting to a depressed quartic. ==Transformation by a rational function==

All preceding examples are polynomial transformations by a rational function, also called Tschirnhaus transformations. Let :f(x)=\frac{g(x)}{h(x)} be a rational function, where and are coprime polynomials. The polynomial transformation of a polynomial by is the polynomial (defined up to the product by a non-zero constant) whose roots are the images by of the roots of . Such a polynomial transformation may be computed as a resultant. In fact, the roots of the desired polynomial are exactly the complex numbers such that there is a complex number such that one has simultaneously (if the coefficients of and are not real or complex numbers, "complex number" has to be replaced by "element of an algebraically closed field containing the coefficients of the input polynomials") :\begin{align} P(x)&=0\\ y\,h(x)-g(x)&=0\,. \end{align} This is exactly the defining property of the resultant :\operatorname{Res}_x(y\,h(x)-g(x),P(x)). This is generally difficult to compute by hand. However, as most computer algebra systems have a built-in function to compute resultants, it is straightforward to compute it with a computer. Properties If the polynomial is irreducible, then either the resulting polynomial is irreducible, or it is a power of an irreducible polynomial. Let \alpha be a root of and consider , the field extension generated by \alpha. The former case means that f(\alpha) is a primitive element of , which has as minimal polynomial. In the latter case, f(\alpha) belongs to a subfield of and its minimal polynomial is the irreducible polynomial that has as power. ==Transformation for equation-solving==

Polynomial transformations have been applied to the simplification of polynomial equations for solution, where possible, by radicals. Descartes introduced the transformation of a polynomial of degree which eliminates the term of degree by a translation of the roots. Such a polynomial is termed depressed. This already suffices to solve the quadratic by square roots. In the case of the cubic, Tschirnhaus transformations replace the variable by a quadratic function, thereby making it possible to eliminate two terms, and so can be used to eliminate the linear term in a depressed cubic to achieve the solution of the cubic by a combination of square and cube roots. The Bring–Jerrard transformation, which is quartic in the variable, brings a quintic into Bring-Jerrard normal form with terms of degree 5,1, and 0. ==See also==