How do you solve #x^3-5x^2<x-5# using a sign chart?

1 Answer
Feb 13, 2017

Answer:

The answer is #x in ]-oo, -1 [uu]1,5[#

Explanation:

Let's rewrite the inequality

#x^3-5x^2 < x-5#

#x^3-5x^2-x+5<0#

Let #f(x)=x^3-5x^2-x+5#

#f(1)=1-5-1+5 =0#

so,

#(x-1)# is a factor of #f(x)#

#f(-1)=-1-5+1+5=0#

so,

#(x+1)# is a factor of #f(x)#

So,

#(x+1)(x-1)=x^2-1# is a factor of #f(x)#

We perform a long division to find the 3rd factor

#color(white)(aaaa)##x^3-5x^2-x+5##color(white)(aaaa)##|##x^2-1#

#color(white)(aaaa)##x^3##color(white)(aaaaaa)##-x##color(white)(aaaaaaaa)##|##x-5#

#color(white)(aaaa)##0-5x^2##color(white)(aaa)##0+5#

#color(white)(aaaaaa)##-5x^2##color(white)(aaaaa)##+5#

#color(white)(aaaaaaaaa)##0##color(white)(aaaaa)##+0#

Therefore,

#f(x)=(x+1)(x-1)(x-5)#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##1##color(white)(aaaaa)##5##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-5##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<0# when #x in ]-oo, -1 [uu]1,5[#