# How do you find all rational roots for 8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0?

Aug 4, 2016

The rational roots are $\frac{1}{4}$, $\frac{1}{2}$.

The remaining two roots are $\pm \sqrt{2} i$

#### Explanation:

$f \left(y\right) = 8 {y}^{4} - 6 {y}^{3} + 17 {y}^{2} - 12 y + 2$

By the rational root theorem, any rational zeros of $f \left(y\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $8$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{8} , \pm \frac{1}{4} , \pm \frac{1}{2} , \pm 1 , \pm 2$

In addition, note that $f \left(- y\right) = 8 {y}^{4} + 6 {y}^{3} + 17 {y}^{2} + 12 y + 2$ has all positive coefficients. Hence $f \left(y\right)$ has no negative zeros.

So the only possible rational zeros of $f \left(y\right)$ are:

$\frac{1}{8} , \frac{1}{4} , \frac{1}{2} , 1 , 2$

We find:

$f \left(\frac{1}{4}\right) = 8 {\left(\frac{1}{4}\right)}^{4} - 6 {\left(\frac{1}{4}\right)}^{3} + 17 {\left(\frac{1}{4}\right)}^{2} - 12 \left(\frac{1}{4}\right) + 2$

$= \frac{1 - 3 + 34 - 96 + 64}{32} = 0$

$f \left(\frac{1}{2}\right) = 8 {\left(\frac{1}{2}\right)}^{4} - 6 {\left(\frac{1}{2}\right)}^{3} + 17 {\left(\frac{1}{2}\right)}^{2} - 12 \left(\frac{1}{2}\right) + 2$

$= \frac{2 - 3 + 17 - 24 + 8}{4} = 0$

So $y = \frac{1}{4}$ and $y = \frac{1}{2}$ are zeros and $\left(4 y - 1\right)$ and $\left(2 y - 1\right)$ are factors:

$8 {y}^{4} - 6 {y}^{3} + 17 {y}^{2} - 12 y + 2$

$= \left(4 y - 1\right) \left(2 {y}^{3} - {y}^{2} + 4 y - 2\right)$

$= \left(4 y - 1\right) \left(2 y - 1\right) \left({y}^{2} + 2\right)$

${y}^{2} + 2 \ge 2 > 0$ for all Real values of $y$, so there are no more Real, let alone rational, zeros.

The last two zeros are $\pm \sqrt{2} i$ since:

$\left(y - \sqrt{2} i\right) \left(y + \sqrt{2} i\right) = {y}^{2} - {\left(\sqrt{2} i\right)}^{2} = {y}^{2} + 2$