How do you solve #2x^3-3x^2-32x+48>=0# using a sign chart?

1 Answer
Jan 15, 2017

Answer:

The answer is #x in [-4,3/2] uu [4, +oo[#

Explanation:

Let #f(x)=2x^3-3x^2-32x+48#

Then

#f(4)=128-48-128+48=0#

So, #(x-4)# is a factor of #f(x)#

To find the other factors, we do a long division

#color(white)(aaaa)##2x^3-3x^2-32x+48##color(white)(aaaa)##∣##color(red)(x-4)#

#color(white)(aaaa)##2x^3-8x^2##color(white)(aaaaaaaaaaaaaa)##∣##color(blue)(2x^2+5x-12)#

#color(white)(aaaaaa)##0+5x^2-32x#

#color(white)(aaaaaaaa)##+5x^2-20x#

#color(white)(aaaaaaaaaa)##+0-12x+48#

#color(white)(aaaaaaaaaaaaaa)##-12x+48#

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

Therefore,

#f(x)=2x^3-3x^2-32x+48=(x-4)(2x^2+5x-12)#

#=(x-4)(2x-3)(x+4)#

Now, we can make the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaaa)##3/2##color(white)(aaaa)##4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##2x-3##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-4##color(white)(aaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaa)##+#

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

Therefore,

#f(x)>=0# when # x in [-4,3/2] uu [4, +oo[ #