As #x(3-x)(x-5)<=0#, we have following options
either
(1) #x<0# - in such a case #x# and #x-5# are negative and #3-x# is positive and hence #x(3-x)(x-5)# is positive and hence #x<0# is not a solution.
(2) #0<=x<=3# - in such a case #x# and #3-x# are positive and #x-5# is negative and hence #x(3-x)(x-5)# is negative and hence #0<=x<=3# is a solution .
(3) #3< x< 5# - in such a case #x-5# and #3-x# are negative and #x# is positive and hence #x(3-x)(x-5)# is positive and hence #3< x< 5# is not a solution .
(4) #x<=5# - in such a case #x# and #x-5# are positive and #3-x# is negative and hence #x(3-x)(x-5)# is negative and hence #0<=x<=3# is a solution.
Hence solution is #0<=x<=3# or #x<=5#.
graph{x(3-x)(x-5) [-10, 10, -10, 10]}