Polynomial transformation

Let

${\displaystyle P(x)=a_{0}x^{n}+a_{1}x^{n-1}+\cdots +a_{n}}${\displaystyle P(x)=a_{0}x^{n}+a_{1}x^{n-1}+\cdots +a_{n}}

be a polynomial, and

${\displaystyle \alpha _{1},\ldots ,\alpha _{n}}${\displaystyle \alpha _{1},\ldots ,\alpha _{n}}

be its complex roots (not necessarily distinct).

For any constant c, the polynomial whose roots are

${\displaystyle \alpha _{1}+c,\ldots ,\alpha _{n}+c}${\displaystyle \alpha _{1}+c,\ldots ,\alpha _{n}+c}

is

${\displaystyle Q(y)=P(y-c)=a_{0}(y-c)^{n}+a_{1}(y-c)^{n-1}+\cdots +a_{n}.}${\displaystyle Q(y)=P(y-c)=a_{0}(y-c)^{n}+a_{1}(y-c)^{n-1}+\cdots +a_{n}.}

If the coefficients of P are integers and the constant ${\displaystyle c={\frac {p}{q}}}${\displaystyle c={\frac {p}{q}}} is a rational number, the coefficients of Q may be not integers, but the polynomial cⁿ Q has integer coefficients and has the same roots as Q.

A special case is when ${\displaystyle c={\frac {a_{1}}{na_{0}}}.}${\displaystyle c={\frac {a_{1}}{na_{0}}}.} The resulting polynomial Q does not have any term in y^{n − 1}.

Let

${\displaystyle P(x)=a_{0}x^{n}+a_{1}x^{n-1}+\cdots +a_{n}}${\displaystyle P(x)=a_{0}x^{n}+a_{1}x^{n-1}+\cdots +a_{n}}

be a polynomial. The polynomial whose roots are the reciprocals of the roots of P is its reciprocal polynomial

${\displaystyle Q(y)=y^{n}P\left({\frac {1}{y}}\right)=a_{n}y^{n}+a_{n-1}y^{n-1}+\cdots +a_{0}.}${\displaystyle Q(y)=y^{n}P\left({\frac {1}{y}}\right)=a_{n}y^{n}+a_{n-1}y^{n-1}+\cdots +a_{0}.}

Let

${\displaystyle P(x)=a_{0}x^{n}+a_{1}x^{n-1}+\cdots +a_{n}}${\displaystyle P(x)=a_{0}x^{n}+a_{1}x^{n-1}+\cdots +a_{n}}

be a polynomial, and c be a non-zero constant. A polynomial whose roots are the product by c of the roots of P is

${\displaystyle Q(y)=c^{n}P\left({\frac {y}{c}}\right)=a_{0}y^{n}+a_{1}cy^{n-1}+\cdots +a_{n}c^{n}.}${\displaystyle Q(y)=c^{n}P\left({\frac {y}{c}}\right)=a_{0}y^{n}+a_{1}cy^{n-1}+\cdots +a_{n}c^{n}.}

The factor cⁿ appears here because, if c and the coefficients of P are integers or belong to some integral domain, the same is true for the coefficients of Q.

In the special case where ${\displaystyle c=a_{0}}${\displaystyle c=a_{0}}, all coefficients of Q are multiple of c, and ${\displaystyle {\frac {Q}{c}}}${\displaystyle {\frac {Q}{c}}} is a monic polynomial, whose coefficients belong to any integral domain containing c and the coefficients of P. 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 ${\displaystyle {\frac {a_{1}}{na_{0}}}}${\displaystyle {\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 n − 1. For examples of this, see Cubic function § Reduction to a depressed cubic or Quartic function § Converting to a depressed quartic.

If the polynomial P is irreducible, then either the resulting polynomial Q is irreducible, or it is a power of an irreducible polynomial. Let ${\displaystyle \alpha }${\displaystyle \alpha } be a root of P and consider L, the field extension generated by ${\displaystyle \alpha }${\displaystyle \alpha }. The former case means that ${\displaystyle f(\alpha )}${\displaystyle f(\alpha )} is a primitive element of L, which has Q as minimal polynomial. In the latter case, ${\displaystyle f(\alpha )}${\displaystyle f(\alpha )} belongs to a subfield of L and its minimal polynomial is the irreducible polynomial that has Q as power.