How do you solve #x(x-1)(x+3) <0#?

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Answer 1 · 2016-11-29T11:15:58

The answer is #x in ] -oo,-3 [ uu] 0,1 [ #

Let #f(x)=x(x-1)(x+3)#

Let's do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##0##color(white)(aaaa)##1##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaa)##-##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##-##color(white)(aaa)##+#

#f(x)<0# when, #x in ] -oo,-3 [ uu] 0,1 [ #

graph{x(x-1)(x+3) [-7.9, 7.9, -3.95, 3.95]}