Let #f(z) = (z+3)(2-z)(1-2z) = (z+3)(2z-1)(z-2)#
Then #f(z) = 0# when #z = -3#, #z = 1/2# and #z = 2#
These three points split the real line into four intervals:
#(-oo, -3)#, #(-3, 1/2)#, #(1/2,2)# and #(2,oo)#
If #z in (-oo, -3)# then
#(z+3) < 0#, #(2z-1) < 0#, #(z-2) < 0# so #f(z) < 0#
If #color(red)(z in (-3, 1/2))# then
#(z+3) > 0#, #(2z-1) < 0#, #(z-2) < 0# so #color(red)(f(z) > 0)#
If #z in (1/2, 2)# then
#(z+3) > 0#, #(2z-1) > 0#, #(z-2) < 0# so #f(z) < 0#
If #color(red)(z in (2, oo))# then
#(z+3) > 0#, #(2z-1) > 0#, #(z-2) > 0# so #color(red)(f(z) > 0)#
So the solution is #z in (-3, 1/2) uu (2, oo)#
graph{ (x+3)(2-x)(1-2x) [-40, 40, -12.24, 27.76]}
