## Is it Possible to Find Value of a Non Symmetric Polynomial?

A symmetric polynomial\[f(\alpha , \beta , \gamma ,...)\]

is a polynomimal such that if the arguments \[\alpha , \beta , \gamma , \...\]

are raeearanged, then the value of \[f\]

remains the same, because rearranges the arguments only changes the order of the terms.Example:

\[f(\alpha , \beta , \gamma )= \alpha^2 + \beta^2 + \gamma^2\]

is a symmetric polynomial because \[f(\alpha , \beta , \gamma )=f(\alpha , \beta , \gamma )=f(\alpha , \beta , \gamma )= \alpha^2 + \beta^2 + \gamma^2\]

.Suppose

\[f(\alpha , \beta )= \alpha - \beta\]

.\[f(\alpha , \beta )\]

is not a symmetric polynomial because \[f(\alpha , \beta )= \alpha - \beta \neq f(\beta , \alpha )= \beta - \alpha\]

.We can however create a symmetric polynomial from

\[f(\alpha , \beta , \gamma )\]

by squaring it.\[(f(\alpha , \beta ))=(\alpha - \beta )^2 = \alpha^2 - 2 \alpha \beta \beta^2 = \beta^2 - 2 \beta \alpha + \alpha^2 = (f(\beta , \alpha ))^2 \]

.Can we then find the value of

\[f(\alpha , \beta ) = \alpha - \beta\]

by finding the square root of \[(f(\alpha , \beta))^2\]

?Suppose the root of the equation

\[z^2- 3z-5\]

are \[\alpha , \: \beta \]

.Since the roots are

\[\alpha , \: \beta\]

, \[(z- \alpha )(z - \beta )= z^2 - (\alpha + \beta ) z + \alpha \beta =0\]

.Comparing with the equation above,

\[\alpha + \beta = 3, \: \alpha \beta = -5\]

. \[(f(\alpha , \beta ))^2 = z^2 - 2 \alpha \beta + \beta ^2 = (\alpha + \beta )^2 - 4 \alpha \beta =3^2-4 \times -5 =29\]

.Then

\[f(\alpha , \beta )= \pm \sqrt{29}\]

.This will only work if the function

\[f\]

has terms such that powers of roots of symmetric, and the only things that stops symmetry are the signs of some terms.The value of

\[\alpha - 3 \beta\]

cannot be found, and neither can the value of \[\alpha + 3 \beta\]