How do you solve #x^3+7x^2-x-7<0#?

1 Answer
Jan 29, 2017

Answer:

The answer is #x in ]-oo, -7 [uu]-1, 1[#

Explanation:

Let #f(x)=x^3+7x^2-x-7#

#f(1)=1+7-1-7=0#

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

To find the other factors, we do a long division

#color(white)(aaaa)##x^3+7x^2-x-7##color(white)(aaaa)##|##color(blue)(x-1)#

#color(white)(aaaa)##x^3-x^2##color(white)(aaaaaaaaaaaa)##|##color(red)(x^2+8x+7)#

#color(white)(aaaaa)##0+8x^2-x#

#color(white)(aaaaaaa)##+8x^2-8x#

#color(white)(aaaaaaaaa)##+0+7x-7#

#color(white)(aaaaaaaaaaaaa)##+7x-7#

#color(white)(aaaaaaaaaaaaaa)##+0-0#

Therefore,

#f(x)=(x-1)(x^2+8x+7)=(x-1)(x+1)(x+7)#

Now, we can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-7##color(white)(aaaa)##-1##color(white)(aaaa)##1##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+7##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

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

Therefore,

#f(x)<0# when #x in ]-oo, -7 [uu]-1, 1[#