# How do you use the rational root theorem to find the roots of 8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0?

Jul 8, 2015

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

has two rational roots $\frac{1}{2}$ and $\frac{1}{4}$ findable by the rational roots theorem.

#### Explanation:

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

By the rational root theorem, any rational roots of $f \left(y\right) = 0$ must be of the form $\frac{p}{q}$ where $p$ and $q$ are integers, in lowest terms such that $p$ is a divisor of the constant term $2$ and $q$ is a divisor of the coefficient, $8$, of the highest order term.

So the possible rational roots are:

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

Notice that $f \left(- x\right) = 8 {x}^{4} + 6 {x}^{3} + 17 {x}^{2} + 12 x + 2$ has all positive coefficients, so no positive value of $x$ is a root of $f \left(- x\right) = 0$. Hence no negative value of $y$ is a root of $f \left(y\right) = 0$.

So the only possible rational roots are:

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

$f \left(2\right) = 8 \cdot 16 - 6 \cdot 8 + 17 \cdot 4 - 12 \cdot 2 + 2$

$= 128 - 48 + 68 - 24 + 2 = 126$

$f \left(1\right) = 8 - 6 + 17 - 12 + 2 = 9$

$f \left(\frac{1}{2}\right) = \frac{8}{16} - \frac{6}{8} + \frac{17}{4} - \frac{12}{2} + 2$

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

$f \left(\frac{1}{4}\right) = \frac{8}{256} - \frac{6}{64} + \frac{17}{16} - \frac{12}{4} + 2$

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

We can tell in advance that $f \left(\frac{1}{8}\right) \ne 0$, since we have already found roots $y = \frac{1}{2}$ and $y = \frac{1}{4}$, which give factors $\left(2 y - 1\right)$ and $\left(4 y - 1\right)$, which together will exhaust the leading coefficient of ${y}^{4}$.

$\left(2 y - 1\right) \left(4 y - 1\right) = 8 {y}^{2} - 6 y + 1$

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

$= \left(8 {y}^{2} - 6 y + 1\right) \left({y}^{2} + 2\right)$

Notice that ${y}^{2} + 2 \ge 2 > 0$ for all $y \in \mathbb{R}$, so there are no more real roots.