How do you solve #2y^3-y^2-2+1<=0# using a sign chart?

1 Answer
Sep 30, 2017

Answer:

The solution is #y in (-oo,-1]uu[1/2,1]#

Explanation:

I assume that #-2# means #-2y#

Before performing the sign chart, we need the roots of the polynomial

Let #f(y)=2y^3-y^2-2y+1#

#f(1)=2-1-2+1=0#

Therefore,

#(y-1)# is a factor of the polynomial

Therefore,

#2y^3-y^2-2y+1=(y-1)(ay^2+by+c)#

#=ay^3+by^2+cy-ay^2-by-c#

Comparing the coefficients

#a=2#

#b-a=-1#, #=>#, #b=a-1=2-1=1#

#c-b=-2#, #=>#, #c=b-2=1-2=-1#

Therefore,

#2y^3-y^2-2y+1=(y-1)(2y^2+y-1)=(y-1)(2y-1)(y+1)#

We can build the sign chart

#color(white)(aaaa)##y##color(white)(aaaaa)##-oo##color(white)(aaaaa)##-1##color(white)(aaaaaaa)##1/2##color(white)(aaaaaa)##1##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##y+1##color(white)(aaaaaa)##-##color(white)(aaa)##0##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##2y-1##color(white)(aaaaa)##-##color(white)(aaa)####color(white)(aaaaa)##-##color(white)(a)##0##color(white)(aa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##y-1##color(white)(aaaaaa)##-##color(white)(aaa)####color(white)(aaaaa)##-##color(white)(aaa)####color(white)(a)##-##color(white)(aa)##0##color(white)(aaa)##+#

#color(white)(aaaa)##f(y)##color(white)(aaaaaaa)##-##color(white)(aaa)##0##color(white)(aaaa)##+##color(white)(a)##0##color(white)(a)##-##color(white)(aaa)##0##color(white)(aaaa)##+#

Therefore,

#f(y)<=0# when #y in (-oo,-1]uu[1/2,1]#

graph{2x^3-x^2-2x+1 [-5.55, 5.55, -2.773, 2.776]}