How do you solve #x^3-6x^2-4x+24>0# using a sign chart?

1 Answer
Dec 31, 2016

Answer:

The answer is #x in ] -2,+2 [ uu ] 6, +oo[#

Explanation:

Let #f(x)=x^3-6x^2-4x+24#

As a polynomial, the domain of #f(x)# is #D_f(x)=RR#

Then, by trial and error,

#f(2)=8-24-8+24=0#

Therefore,

#(x-2)# is a factor

To find the other factors, we do a long division

#color(white)(aaaa)##x^3-6x^2-4x+24##color(white)(aaaa)##∣##x-2#

#color(white)(aaaa)##x^3-2x^2##color(white)(aaaaaaaaaaaaa)##∣##x^2-4x-12#

#color(white)(aaaa)##0-4x^2-4x#

#color(white)(aaaaaa)##-4x^2+8x#

#color(white)(aaaaaaa)##-0-12x+24#

#color(white)(aaaaaaaaaa)##-12x+24#

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

Therefore,

#x^2-4x-12=(x+2)(x-6)#

and #x^3-6x^2-4x+24=(x-2)(x+2)(x-6)#

Now, we can establish the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaa)##2##color(white)(aaaaa)##6##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-6##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaa)##+#

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

Therefore,

#f(x)>0# when #x in ] -2,+2 [ uu ] 6, +oo[ #