# How do you use the rational roots theorem to find all possible zeros of h(x) = 2x^4-5x^3+3x^2+4x-6?

Jun 24, 2016

Zeros: $x = - 1$, $x = \frac{3}{2}$ and $x = 1 \pm i$

#### Explanation:

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

By the rational roots theorem, any rational zeros of $h \left(x\right)$ are expressible in the form $p , q$ for integers $p , q$ with $p$ 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:

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

Trying each in turn, we first find:

$f \left(- 1\right) = 2 + 5 + 3 - 4 - 6 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

$2 {x}^{4} - 5 {x}^{3} + 3 {x}^{2} + 4 x - 6$

$= \left(x + 1\right) \left(2 {x}^{3} - 7 {x}^{2} + 10 x - 6\right)$

Then substituting $x = \frac{3}{2}$ in the remaining cubic, we find:

$2 {x}^{3} - 7 {x}^{2} + 10 x - 6 = 2 \left(\frac{27}{8}\right) - 7 \left(\frac{9}{4}\right) + 10 \left(\frac{3}{2}\right) - 6$

$= \frac{27}{4} - \frac{63}{4} + \frac{60}{4} - \frac{24}{4} = 0$

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

$2 {x}^{3} - 7 {x}^{2} + 10 x - 6 = \left(2 x - 3\right) \left({x}^{2} - 2 x + 2\right)$

The remaining quadratic has negative discriminant, but we can find its Complex zeros by completing the square:

${x}^{2} - 2 x + 2 = {x}^{2} - 2 x + 1 + 1$

$= {\left(x - 1\right)}^{2} - {i}^{2}$

$= \left(x - 1 - i\right) \left(x - 1 + i\right)$

Hence zeros $x = 1 \pm i$