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

1 Answer
Sep 30, 2017

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)ycolor(white)(aaaaa)-oocolor(white)(aaaaa)-1color(white)(aaaaaaa)1/2color(white)(aaaaaa)1color(white)(aaaaaa)+oo

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

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

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

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