# How do you solve (x+5)(x-2)(x-1)(x+1)<0?

Aug 12, 2016

$x \in \left(- 5 , - 1\right) \cup \left(1 , 2\right)$

#### Explanation:

Quick method

The left hand side is an upright quartic (positive coefficient of ${x}^{4}$) which intersects the $x$ axis at the four zeros: $- 5 , - 1 , 1 , 2$.

So it will be negative precisely in the intervals $\left(- 5 , - 1\right)$ and $\left(1 , 2\right)$.

Here's a graph of $y = \frac{1}{20} \left(x + 5\right) \left(x - 2\right) \left(x - 1\right) \left(x + 1\right)$ ...

graph{(x+5)(x+1)(x-1)(x-2)/20 [-10, 10, -5.5, 4.5]}

$\textcolor{w h i t e}{}$
A little slower...

Let $f \left(x\right) = \left(x + 5\right) \left(x - 2\right) \left(x - 1\right) \left(x + 1\right)$

$f \left(x\right)$ is a continuous function which is zero when $x \in \left\{- 5 , - 1 , 1 , 2\right\}$, so does not satisfy the inequality $f \left(x\right) < 0$ at those points.

These points divide the Real line into $5$ intervals in which the sign of $f \left(x\right)$ cannot change:

$\left(- \infty , - 5\right) , \left(- 5 , - 1\right) , \left(- 1 , 1\right) , \left(1 , 2\right) , \left(2 , \infty\right)$

Since the zeros are each of multiplicity $1$ - which is an odd number - the sign does change at each of these points.

When $x < - 5$, the sign of each of the linear factors is negative. When they are all multiplied, the result is positive.

That is, when $x < - 5$:

$\left\{\begin{matrix}\left(x + 5\right) < 0 \\ \left(x - 2\right) < 0 \\ \left(x - 1\right) < 0 \\ \left(x + 1\right) < 0\end{matrix}\right.$

Hence $\left(x + 5\right) \left(x - 2\right) \left(x - 1\right) \left(x + 1\right) > 0$

So $f \left(x\right)$ has the following signs in each of the $5$ intervals:

$+ , - , + , - , +$

respectively.

So $f \left(x\right) < 0$ in $\left(- 5 , - 1\right) \cup \left(1 , 2\right)$