How do you solve #2x^3+5x^2>6x+9# using a sign chart?

1 Answer
Dec 30, 2016

Answer:

The answer is #x in ] -3, -1 [ uu ]3/2, +oo[#

Explanation:

Let's rewrite the inequality as

#2x^3+5x^2-6x-9>0#

We must factorise the LHS

Let #f(x)=2x^3+5x^2-6x-9#

We calculate

#f(-1)=-2+5+6-9=0#

Therefore, #(x+1)# is a factor

To find the other factors, we do a long division

#color(white)(aaaa)##2x^3+5x^2-6x-9##color(white)(aaaa)##∣##x+1#

#color(white)(aaaa)##2x^3+2x^2##color(white)(aaaaaaaaaaaa)##∣##2x^2+3x-9#

#color(white)(aaaaaa)##0+3x^2-6x#

#color(white)(aaaaaaaa)##+3x^2+3x#

#color(white)(aaaaaaaaaa)##+0-9x-9#

#color(white)(aaaaaaaaaaaaaa)##-9x-9#

#color(white)(aaaaaaaaaaaaaaa)##-0-0#

Therefore,

#(2x^3+5x^2-6x-9)/(x+1)=2x^2+3x-9=(2x-3)(x+3)#

So,

#2x^3+5x^2-6x-9=(x+1)(2x-3)(x+3)>0#

Now we can do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-3##color(white)(aaaa)##-1##color(white)(aaaa)##3/2##color(white)(aaaa)##+oo#

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

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

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

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

Therefore,

#f(x)>0# when #x in ] -3, -1 [ uu ]3/2, +oo[#