How do you use the rational roots theorem to find all possible zeros of f(x)=2x^3+x^2-13x+6?

Mar 23, 2016

$2 {x}^{3} + {x}^{2} - 13 x + 6 = \left(2 x - 1\right) \left(x + 3\right) \left(x - 2\right)$

Explanation:

$f \left(x\right) = 2 {x}^{3} + {x}^{2} - 13 x + 6$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$ and $q$ where $p$ is a divisor of the constant term $6$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2}$, $\pm 1$, $\pm \frac{3}{2}$, $\pm 2$, $\pm 3$, $\pm 6$

Let us try each in turn:

$f \left(\frac{1}{2}\right) = \frac{1}{4} + \frac{1}{4} - \frac{13}{2} + 6 = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$2 {x}^{3} + {x}^{2} - 13 x + 6 = \left(2 x - 1\right) \left({x}^{2} + x - 6\right)$

We could continue simply trying the other possible zeros, but it is quicker to note that $3 \times 2 = 6$ and $3 - 2 = 1$, so:

${x}^{2} + x - 6 = \left(x + 3\right) \left(x - 2\right)$

Putting it all together, we find:

$2 {x}^{3} + {x}^{2} - 13 x + 6 = \left(2 x - 1\right) \left(x + 3\right) \left(x - 2\right)$