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

Sep 30, 2017

The solution is $y \in \left(- \infty , - 1\right] \cup \left[\frac{1}{2} , 1\right]$

#### Explanation:

I assume that $- 2$ means $- 2 y$

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

Let $f \left(y\right) = 2 {y}^{3} - {y}^{2} - 2 y + 1$

$f \left(1\right) = 2 - 1 - 2 + 1 = 0$

Therefore,

$\left(y - 1\right)$ is a factor of the polynomial

Therefore,

$2 {y}^{3} - {y}^{2} - 2 y + 1 = \left(y - 1\right) \left(a {y}^{2} + b y + c\right)$

$= a {y}^{3} + b {y}^{2} + c y - a {y}^{2} - b y - c$

Comparing the coefficients

$a = 2$

$b - a = - 1$, $\implies$, $b = a - 1 = 2 - 1 = 1$

$c - b = - 2$, $\implies$, $c = b - 2 = 1 - 2 = - 1$

Therefore,

$2 {y}^{3} - {y}^{2} - 2 y + 1 = \left(y - 1\right) \left(2 {y}^{2} + y - 1\right) = \left(y - 1\right) \left(2 y - 1\right) \left(y + 1\right)$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a a a}$$\frac{1}{2}$$\textcolor{w h i t e}{a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$y + 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$2 y - 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$y - 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a a a}$color(white)(a)-$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(y\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(y\right) \le 0$ when $y \in \left(- \infty , - 1\right] \cup \left[\frac{1}{2} , 1\right]$

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