Question

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

Answer 1

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]}