Convert to sqrfree performs a square-free factorization of the polynomial a.  It is most often called with the third argument x specified.  In this case, content( a, x) is first removed (thus making a primitive) before doing the square-free factorization.  The resulting factorization is of the form:

If called with only two arguments, a complete square-free factorization will be done.  A variable is first chosen for which the above computation is done.  Next, the content is made square-free by recursive application of convert to sqrfree.  This is repeated until there are no variables left in the content.

It is important to realize that the factors will be in random order, not in order of increasing multiplicity as may be expected.  Also, the multiplicities are not necessarily distinct as partial factorizations in the input are preserved as much as possible.

$\mathrm{convert}\left(x^2+4x+4\,\mathrm{sqrfree}\,x\right)$

${\left(x+2\right)}^2$

$\mathrm{poly}≔y^2{x}^3+2y^2{x}^2+y^{2}x+2y{x}^3+4y{x}^2+2yx\:$

$\mathrm{convert}\left(\mathrm{poly}\,\mathrm{sqrfree}\,x\right)$

$y\left(y+2\right)x{\left(x+1\right)}^2$

$\mathrm{convert}\left(\mathrm{poly}\,\mathrm{sqrfree}\right)$

$y\left(y+2\right)x{\left(x+1\right)}^2$

$g≔\left(x+y+1\right)\mathrm{expand}\left({\left(x+y+1\right)}^2\right){\left(x-y-3\right)}^3\left(3x+6y-21\right)$

$g≔\left(x+y+1\right)\left(x^2+2yx+y^2+2x+2y+1\right){\left(x-y-3\right)}^3\left(3x+6y-21\right)$

$\mathrm{convert}\left(g\,\mathrm{sqrfree}\,\left[x\,y\right]\right)$

$3{\left(x+y+1\right)}^3{\left(x-y-3\right)}^3\left(x+2y-7\right)$