# Solve (z+3)(2-z)(1-2z)>0?

Jul 10, 2015

$z \in \left(- 3 , \frac{1}{2}\right) \cup \left(2 , \infty\right)$

#### Explanation:

Let $f \left(z\right) = \left(z + 3\right) \left(2 - z\right) \left(1 - 2 z\right) = \left(z + 3\right) \left(2 z - 1\right) \left(z - 2\right)$

Then $f \left(z\right) = 0$ when $z = - 3$, $z = \frac{1}{2}$ and $z = 2$

These three points split the real line into four intervals:

$\left(- \infty , - 3\right)$, $\left(- 3 , \frac{1}{2}\right)$, $\left(\frac{1}{2} , 2\right)$ and $\left(2 , \infty\right)$

If $z \in \left(- \infty , - 3\right)$ then

$\left(z + 3\right) < 0$, $\left(2 z - 1\right) < 0$, $\left(z - 2\right) < 0$ so $f \left(z\right) < 0$

If $\textcolor{red}{z \in \left(- 3 , \frac{1}{2}\right)}$ then

$\left(z + 3\right) > 0$, $\left(2 z - 1\right) < 0$, $\left(z - 2\right) < 0$ so $\textcolor{red}{f \left(z\right) > 0}$

If $z \in \left(\frac{1}{2} , 2\right)$ then

$\left(z + 3\right) > 0$, $\left(2 z - 1\right) > 0$, $\left(z - 2\right) < 0$ so $f \left(z\right) < 0$

If $\textcolor{red}{z \in \left(2 , \infty\right)}$ then

$\left(z + 3\right) > 0$, $\left(2 z - 1\right) > 0$, $\left(z - 2\right) > 0$ so $\textcolor{red}{f \left(z\right) > 0}$

So the solution is $z \in \left(- 3 , \frac{1}{2}\right) \cup \left(2 , \infty\right)$

graph{ (x+3)(2-x)(1-2x) [-40, 40, -12.24, 27.76]}