# What are the possible rational roots of 2x^3+3x^2-8x+3=0 and then determine the rational roots?

Jan 31, 2017

The "possible" rational roots are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 3$

The actual roots are:

$1 , \frac{1}{2} , - 3$

#### Explanation:

Given:

$2 {x}^{3} + 3 {x}^{2} - 8 x + 3 = 0$

By the rational roots theorem, any rational zeros of this cubic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $3$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 3$

In addition, note that the sum of the coefficients is $0$. That is:

$2 + 3 - 8 + 3 = 0$

Hence $x = 1$ is a root and $\left(x - 1\right)$ a factor:

$0 = 2 {x}^{3} + 3 {x}^{2} - 8 x + 3$

$\textcolor{w h i t e}{0} = \left(x - 1\right) \left(2 {x}^{2} + 5 x - 3\right)$

To find the zeros of the remaining quadratic we could try each of the other possible rational roots in turn, but I'd rather complete the square:

$0 = 8 \left(2 {x}^{2} + 5 x - 3\right)$

$\textcolor{w h i t e}{0} = 16 {x}^{2} + 40 x - 24$

$\textcolor{w h i t e}{0} = {\left(4 x\right)}^{2} + 2 \left(4 x\right) \left(5\right) + {\left(5\right)}^{2} - 49$

$\textcolor{w h i t e}{0} = {\left(4 x + 5\right)}^{2} - {7}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(4 x + 5\right) - 7\right) \left(\left(4 x + 5\right) + 7\right)$

$\textcolor{w h i t e}{0} = \left(4 x - 2\right) \left(4 x + 12\right)$

$\textcolor{w h i t e}{0} = \left(2 \left(2 x - 1\right)\right) \left(4 \left(x + 3\right)\right)$

$\textcolor{w h i t e}{0} = 8 \left(2 x - 1\right) \left(x + 3\right)$

So the remaining two roots are:

$x = \frac{1}{2} \text{ }$ and $\text{ } x = - 3$