# How do you solve y^3-y^2-4y+4>0 using a sign chart?

Sep 20, 2016

Solution is either $- 2 < y < 1$ or $y > 2$

#### Explanation:

Let us first factorize ${y}^{3} - {y}^{2} - 4 y + 4$.

${y}^{3} - {y}^{2} - 4 y + 4 = {y}^{2} \left(y - 1\right) - 4 \left(y - 1\right) = \left({y}^{2} - 4\right) \left(y - 1\right) = \left(y + 2\right) \left(y - 2\right) \left(y - 1\right)$

Hence we have to solve the inequality ${y}^{3} - {y}^{2} - 4 y + 4 > 0$ or $\left(y + 2\right) \left(y - 2\right) \left(y - 1\right) > 0$

From this we know that the product $\left(y + 2\right) \left(y - 2\right) \left(y - 1\right)$ is positive. It is apparent that sign of binomials $\left(y + 2\right)$, $\left(y - 2\right)$ and $\left(y - 1\right)$ will change around the values $- 2$. $2$ and $1$ respectively. In sign chart we divide the real number line using (y+2)(y-2)(y-1)these values, i.e. below $- 2$, between $- 2$ and $1$, between $1$ and $2$ and above $2$ and see how the sign of $\left(y + 2\right) \left(y - 2\right) \left(y - 1\right)$ changes.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} - 2 \textcolor{w h i t e}{X X X X X} 1 \textcolor{w h i t e}{X X X X X} 2$

$\left(y + 2\right) \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} + i v e$

$\left(y - 2\right) \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e$

$\left(y - 1\right) \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} + i v e$

$\left(y + 2\right) \left(y - 2\right) \left(y - 1\right)$
$\textcolor{w h i t e}{X X X X X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e$

It is observed that $\left(y + 2\right) \left(y - 2\right) \left(y - 1\right) > 0$ when either $- 2 < y < 1$ or $y > 2$, which is the solution for the inequality.