# square root of polynomial

The   $f$,  denoted by $\sqrt{f}$, is any polynomial $g$ having the square  $g^{2}$ equal to $f$.  For example,  $\sqrt{9x^{2}\!-\!30x\!+\!25}=3x\!-\!5$ or $-3x\!+\!5$.

A polynomial needs not have a square root, but if it has a square root $g$, then also the opposite polynomial $-g$ is its square root.

The idea of the squaring

 $(a\!+\!b\!+\!c+..)^{2}=(a)a+(2a\!+\!b)b+(2a\!+\!2b\!+\!c)c+..$

(see the square of sum) gives a method for getting the square root of a polynomial:

• The .

• And so on.

In the examples below, on the .

Example 1.$\sqrt{9x^{4}\!+\!6x^{3}\!-\!11x^{2}\!-\!4x\!+\!4}=$ ?

 $\sqrt{}$ $(9x^{4}$ $+6x^{3}$ $-11x^{2}$ $-4x$ $+4)$ $=$ $\pm$ $(3x^{2}$ $+x$ $-2)$ $9x^{4}$ $3x^{2}$ $6x^{3}$ $-11x^{2}$ $6x^{2}$ $+x$ $6x^{3}$ $+x^{2}$ $x$ $-12x^{2}$ $-4x$ $+4$ $6x^{2}$ $+2x$ $-2$ $-12x^{2}$ $-4x$ $+4$ $-2$ $0$

Example 2.$\sqrt{x^{6}\!-\!2x^{5}\!-\!x^{4}\!+\!3x^{2}\!+\!2x\!+\!1}=$ ?

 $\sqrt{}$ $(1$ $+2x$ $+3x^{2}$ $-x^{4}$ $-2x^{5}$ $+x^{6})$ $=$ $\pm$ $(1$ $+x$ $+x^{2}$ $-x^{3})$ $1$ $1$ $2x$ $+3x^{2}$ $2$ $+x$ $2x$ $+x^{2}$ $+x$ $2x^{2}$ $-x^{4}$ $2$ $+2x$ $+x^{2}$ $2x^{2}$ $+2x^{3}$ $+x^{4}$ $x^{2}$ $-2x^{3}$ $-2x^{4}$ $-2x^{5}$ $+x^{6}$ $2$ $+2x$ $+2x^{2}$ $-x^{3}$ $-2x^{3}$ $-2x^{4}$ $-2x^{5}$ $+x^{6}$ $-x^{3}$ $0$

Remark.  The procedure may give a Taylor series expansion of the square root, if it is not a polynomial.  E.g. we get

 $\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^{2}+\frac{1}{16}x^{3}-\frac{5}{128}x^{4% }+-...$

## References

• 1 Meyers Rechenduden.  Erster verbesserter Neudruck.  Bibliographisches Institut AG, Mannheim (1960).
Title square root of polynomial SquareRootOfPolynomial 2013-03-22 15:32:06 2013-03-22 15:32:06 pahio (2872) pahio (2872) 17 pahio (2872) Algorithm msc 26C99 msc 12E05 calculation of square root of polynomial SquareOfSum BombellisMethodOfComputingSquareRoots