Question 552935
Dividing is multiplying by the inverse;
{{{(2x^2+x-6)/(x^2+4x-5)}}}/{{{(4x^2-6x)/(x^3-3x^2+2x)}}} = {{{((2x^2+x-6)/(x^2+4x-5))((x^3-3x^2+2x)/(4x^2-6x))}}} = {{{((2x^2+x-6)(x^3-3x^2+2x))/((x^2+4x-5)(4x^2-6x))}}}
Factoring, you can simplify
{{{((2x^2+x-6)(x^3-3x^2+2x))/((x^2+4x-5)(4x^2-6x))}}} = {{{((2x-3)(x+2)(x)(x-2)(x-1))/((x+5)(x-1)(2x)(2x-3))}}} = {{{(((x+2)(x)(x-2))/((x+5)(2x)))((x-1)/(x-1))((2x-3)/(2x-3))}}} = {{{(x^3-4x)/(2x^2+10x)}}}