# How do you solve 2x^3 + x^2 - 5x + 2 < 0?

Oct 1, 2015

$2 {x}^{3} + {x}^{2} - 5 x + 2 = \left(x + 2\right) \left(2 x - 1\right) \left(x - 1\right)$

Hence $2 {x}^{3} + {x}^{2} - 5 x + 2 < 0$ when $x \in \left(- \infty , - 2\right) \cup \left(\frac{1}{2} , 1\right)$

#### Explanation:

Let $f \left(x\right) = 2 {x}^{3} + {x}^{2} - 5 x + 2$.

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $q \ne 0$, $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational roots are $\pm \frac{1}{2}$, $\pm 1$ and $\pm 2$.

We find $f \left(- 2\right) = f \left(\frac{1}{2}\right) = f \left(1\right) = 0$, so $- 2$, $\frac{1}{2}$ and $1$ are the three roots.

$f \left(x\right)$ can potentially change sign at each of these roots, and will do since none of them is a repeated root.

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

So when $x < - 2$, the signs of the three factors are $-$, $-$ and $-$, so their product $f \left(x\right) < 0$.

When $- 2 < x < \frac{1}{2}$, the signs of the three factors are $+$, $-$ and $-$, so $f \left(x\right) > 0$.

When $\frac{1}{2} < x < 1$, the signs of the factors are $+$, $+$ and $-$, so $f \left(x\right) < 0$.

When $1 < x$, the signs of the factors are $+$, $+$ and $+$, so $f \left(x\right) > 0$.

So we find $f \left(x\right) < 0$ when $x \in \left(- \infty , - 2\right) \cup \left(\frac{1}{2} , 1\right)$