How do you solve #m^3-2m^2-15m>0# using a sign chart?

Question

1285 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-06T16:01:33

The solution is #m in ]-3,0 [uu]5, +oo[#

Let's factorise the inequality

#m^3-2m^2-15m>0#

#m(m^2-2m-15)>0#

#m(m+3)(m-5)>0#

Let #f(m)=m(m+3)(m-5)#

Now, we can build the sign chart

#color(white)(aaaa)##m##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##0##color(white)(aaaaa)##5##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##m+3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##m##color(white)(aaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##m-5##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(m)##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(m)>0#, when #m in ]-3,0 [uu]5, +oo[#