# What are all the possible rational zeros for f(x)=2x^3+x^2+8x-21 and how do you find all zeros?

Sep 9, 2016

$f \left(x\right)$ has rational zero $x = \frac{3}{2}$ and other zeros $x = - 1 \pm \sqrt{6} i$

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + {x}^{2} + 8 x - 21$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 21$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 3 , \pm \frac{7}{2} , \pm 7 , \pm \frac{21}{2} , \pm 21$

In addition note that the signs of the coefficients are in the pattern $+ + + -$. By Descartes' rule of signs, since there is one change of sign, there is exactly one positive Real zero.

So let's try the positive rational possibilities first:

$f \left(\frac{1}{2}\right) = 2 \left(\frac{1}{8}\right) + \left(\frac{1}{4}\right) + 8 \left(\frac{1}{2}\right) - 21 = - \frac{33}{2}$

$f \left(1\right) = 2 + 1 + 8 - 21 = - 10$

$f \left(\frac{3}{2}\right) = 2 \left(\frac{27}{8}\right) + \left(\frac{9}{4}\right) + 8 \left(\frac{3}{2}\right) - 21 = \frac{27 + 9 + 48 - 84}{4} = 0$

So $x = \frac{3}{2}$ is a zero and $\left(2 x - 3\right)$ a factor:

$2 {x}^{3} + {x}^{2} + 8 x - 21 = \left(2 x - 3\right) \left({x}^{2} + 2 x + 7\right)$

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x + 1\right)$ and $b = \sqrt{6} i$ as follows:

${x}^{2} + 2 x + 7 = {x}^{2} + 2 x + 1 + 6$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 7} = {\left(x + 1\right)}^{2} - {\left(\sqrt{6} i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 7} = \left(\left(x + 1\right) - \sqrt{6} i\right) \left(\left(x + 1\right) + \sqrt{6} i\right)$

$\textcolor{w h i t e}{{x}^{2} + 2 x + 7} = \left(x + 1 - \sqrt{6} i\right) \left(x + 1 + \sqrt{6} i\right)$

Hence zeros:

$x = - 1 \pm \sqrt{6} i$