Solve (z+3)(2-z)(1-2z)>0?

1 Answer
Jul 10, 2015

#z in (-3, 1/2) uu (2, oo)#

Explanation:

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]}