We are solving the inequality #x(x-1)(x+3)>0#. From this we know that the product #x(x-1)(x+3)# is positive. It is also apparent that sign of binomials #(x+3)#, #x# and #(x-1)# will change around the values #-3#. #9# and #1# respectively. We divide the real number line using around these values, i.e. below #-3#, between #-3# and #0#, between #0# and #1# and above #1# and see how the sign of ## changes, as we move across he real number line.

**Sign Chart**

#color(white)(XXXXXXXXXXX)-3color(white)(XXXXX)0color(white)(XXXXX)1#

#(x+3)color(white)(XXXX)-ive color(white)(XXXX)+ive color(white)(XX)+ive color(white)(XXX)+ive#

#xcolor(white)(XXXXXXX)-ive color(white)(XXXX)-ive color(white)(XX)+ive color(white)(XXX)+ive#

#(x-1)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XX)-ive color(white)(XXX)+ive#

#x(x-1)(x+3)#

#color(white)(XXXXXXXX)-ive color(white)(XXXX)+ive color(white)(XX)-ive color(white)(XXX)+ive#

It is observed that #x(x-1)(x+3) > 0# when either #-3 < x < 0# or #x > 1#, which is the solution for the inequality.

This can be written in interval notation as #(-3,0)# or #(1,oo)#.

Here small brackets #()# indicate that endpoints are not included though solution includes values in the interval. This is used when we have only inequality..

If we have equality included say for #x(x-1)(x+3) >= 0#, we use square brackets, which means endpoints are included.