jmc

We can write $$a{x}^4+b{x}^3+32x^2-16x+6=(3x^2-2x+1)(c{x}^2+dx+6).$$Expanding, we get $$a{x}^2+b{x}^3+32x^2-16x+6=3c{x}^4+(-2c+3d)x^3+(c-2d+18)x^2+(d-12)x+6.$$Comparing coefficients, we get $$\begin{array}{rl}a& =3c,\\ b& =-2c+3d,\\ 32& =c-2d+18,\\ -16& =d-12.\end{array}$$Solving, we find $a=18,$ $b=-24,$ $c=6,$ and $d=-4,$ so $(a,b)=\boxed{(18,-24)}.$