# What are all the possible rational zeros for f(x)=3x^5-6x^4+2x^2-6x+12 and how do you find all zeros?

Sep 1, 2017

Assuming:

$f \left(x\right) = 3 {x}^{5} - 6 {x}^{4} - {x}^{3} + 2 {x}^{2} - 6 x + 12$

Possible rational zeros:

$\pm \frac{1}{3} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

Actual zeros:

$2 , \text{ "+-sqrt(1/6(1+sqrt(73)))," } \pm \sqrt{\frac{1}{6} \left(\sqrt{73} - 1\right)} i$

#### Explanation:

I think there is a missing term $- {x}^{3}$ in the question as:

• The given quintic has no solutions expressible using radicals, let alone rational roots.

• There is no ${x}^{3}$ term - so one might well have been missed.

• If the missing term is $- {x}^{3}$ then the quintic is solvable using precalculus level methods and has one rational zero.

So, suppose:

$f \left(x\right) = 3 {x}^{5} - 6 {x}^{4} - {x}^{3} + 2 {x}^{2} - 6 x + 12$

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 $12$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

We find:

$f \left(2\right) = 3 {\left(\textcolor{b l u e}{2}\right)}^{5} - 6 {\left(\textcolor{b l u e}{2}\right)}^{4} - {\left(\textcolor{b l u e}{2}\right)}^{3} + 2 {\left(\textcolor{b l u e}{2}\right)}^{2} - 6 \left(\textcolor{b l u e}{2}\right) + 12$

$\textcolor{w h i t e}{f \left(2\right)} = 96 - 96 - 8 + 8 - 12 + 12 = 0$

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

$3 {x}^{5} - 6 {x}^{4} - {x}^{3} + 2 {x}^{2} - 6 x + 12 = \left(x - 2\right) \left(3 {x}^{4} - {x}^{2} - 6\right)$

Then:

$12 \left(3 {x}^{4} - {x}^{2} - 6\right) = 36 {x}^{4} - 12 {x}^{2} - 72$

$\textcolor{w h i t e}{12 \left(3 {x}^{4} - {x}^{2} - 6\right)} = {\left(6 {x}^{2}\right)}^{2} - 2 \left(6 {x}^{2}\right) + 1 - 73$

$\textcolor{w h i t e}{12 \left(3 {x}^{4} - {x}^{2} - 6\right)} = {\left(6 {x}^{2} - 1\right)}^{2} - {\left(\sqrt{73}\right)}^{2}$

$\textcolor{w h i t e}{12 \left(3 {x}^{4} - {x}^{2} - 6\right)} = \left(\left(6 {x}^{2} - 1\right) - \sqrt{73}\right) \left(\left(6 {x}^{2} - 1\right) + \sqrt{73}\right)$

$\textcolor{w h i t e}{12 \left(3 {x}^{4} - {x}^{2} - 6\right)} = \left(6 {x}^{2} - 1 - \sqrt{73}\right) \left(6 {x}^{2} - 1 + \sqrt{73}\right)$

Hence:

$x = \pm \sqrt{\frac{1}{6} \left(1 + \sqrt{73}\right)} \text{ }$ or $\text{ } x = \pm \sqrt{\frac{1}{6} \left(\sqrt{73} - 1\right)} i$