How do you solve 2x^3+x^2-5x+2<=0 using a sign chart?

Feb 9, 2017

The answer is x in ]-oo, 2] uu [1/2,1]

Explanation:

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

$f \left(1\right) = 2 + 1 - 5 + 2 = 0$

Therefore, $\left(x - 1\right)$ is a factor of $f \left(x\right)$

To find the other factors, we do a long division

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} + {x}^{2} - 5 x + 2$$\textcolor{w h i t e}{a a a a}$$|$$x - 1$

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a}$$|$$2 {x}^{2} + 3 x - 2$

$\textcolor{w h i t e}{a a a a a}$$0 + 3 {x}^{2} - 5 x$

$\textcolor{w h i t e}{a a a a a a a}$$+ 3 {x}^{2} - 3 x$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 0 - 2 x + 2$

$\textcolor{w h i t e}{a a a a a a a a a a a a}$$- 2 x + 2$

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

Therefore,

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

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

Now, we build the sign chart

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

$\textcolor{w h i t e}{a a a a}$$2 x - 1$$\textcolor{w h i t e}{a a a a a}$$-$$\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}$$+$

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

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

Therefore,

$f \left(x\right) \le 0$ when x in ]-oo, 2] uu [1/2,1]