Start by writing out your initial expression
#E = (2x^3 + 3x^2 - 2x)/(3x-15) * (x^2 - 3x - 10)/(2x^3 - x^2) * (5x^2 - 10x)/(3x^2 + 12x + 12)#
You can factor the quadratics by using the sum/product technique to get
#x^2 - 3x - 10 = x^2 + 2x - 5x -10 = (x-5)(x+2)#
and
#3(x^2 + 4x + 4) = 3(x^2 + 2x + 2x + 4) = 3(x+2)(x+2)#
Your expression can thus be written as
#E = (color(blue)(cancel(color(black)(x)))(2x^2 + 3x -2))/(3color(red)(cancel(color(black)((x-5))))) * (color(red)(cancel(color(black)((x-5))))color(purple)(cancel(color(black)((x+2)))))/(x^(color(blue)(cancel(color(black)(2))))(2x-1)) * (5color(blue)(cancel(color(black)(x)))(x-2))/(3color(purple)(cancel(color(black)((x+2))))(x+2))#
#E = (2x^2 + 3x - 2)/3 * 1/(2x-1) * (5(x-2))/(3(x+2))#
The remaining quadratic can be factored by using the quadratic formula
#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))#
#x_(1,2) = (-3 +- sqrt(3^2 - 4 * 2 * (-2)))/(2 *2)#
#x_(1,2) = (-3 +- 5)/4 = { (x_1 = (-3-5)/4 = -2), (x_2 = (-3 + 5)/4 = 1/2) :}#
This means that you have
#E = (color(red)(cancel(color(black)((x+2))))(x-1/2))/3 * 1/(2x-1) * (5(x-2))/(3color(red)(cancel(color(black)((x+2)))))#
#E = 5/9 * ((x-2)color(red)(cancel(color(black)((x-1/2)))))/(2color(red)(cancel(color(black)((x-1/2))))) = color(green)(5/18 * (x-2)#
