# What are the possible rational zeros of P(x) = 4x^4+19x^3-x^2+19x-5 ?

Nov 27, 2017

The "possible" rational zeros are: $\pm \frac{1}{4} , \pm \frac{1}{2} , \pm 1 , \pm \frac{5}{4} , \pm \frac{5}{2} , \pm 5$

The actual rational zeros are: $\frac{1}{4}$ and $- 5$

The other two zeros are: $\pm i$

#### Explanation:

Given:

$P \left(x\right) = 4 {x}^{4} + 19 {x}^{3} - {x}^{2} + 19 x - 5$

By the rational roots theorem, any rational zeros of $P \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 5$ and $q$ a divisor of the coefficient $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm 1 , \pm \frac{5}{4} , \pm \frac{5}{2} , \pm 5$

We find:

$P \left(\frac{1}{4}\right) = 4 {\left(\textcolor{b l u e}{\frac{1}{4}}\right)}^{4} + 19 {\left(\textcolor{b l u e}{\frac{1}{4}}\right)}^{3} - {\left(\textcolor{b l u e}{\frac{1}{4}}\right)}^{2} + 19 \left(\textcolor{b l u e}{\frac{1}{4}}\right) - 5$

$= \frac{1}{64} + \frac{19}{64} - \frac{4}{64} + \frac{304}{64} - \frac{320}{64} = 0$

So $x = \frac{1}{4}$ is a zero and $\left(4 x - 1\right)$ a factor:

$4 {x}^{4} + 19 {x}^{3} - {x}^{2} + 19 x - 5 = \left(4 x - 1\right) \left({x}^{3} + 5 {x}^{2} + x + 5\right)$

$\textcolor{w h i t e}{4 {x}^{4} + 19 {x}^{3} - {x}^{2} + 19 x - 5} = \left(4 x - 1\right) \left(\left({x}^{3} + 5 {x}^{2}\right) + \left(x + 5\right)\right)$

$\textcolor{w h i t e}{4 {x}^{4} + 19 {x}^{3} - {x}^{2} + 19 x - 5} = \left(4 x - 1\right) \left({x}^{2} \left(x + 5\right) + 1 \left(x + 5\right)\right)$

$\textcolor{w h i t e}{4 {x}^{4} + 19 {x}^{3} - {x}^{2} + 19 x - 5} = \left(4 x - 1\right) \left({x}^{2} + 1\right) \left(x + 5\right)$

$\textcolor{w h i t e}{4 {x}^{4} + 19 {x}^{3} - {x}^{2} + 19 x - 5} = \left(4 x - 1\right) \left({x}^{2} - {i}^{2}\right) \left(x + 5\right)$

$\textcolor{w h i t e}{4 {x}^{4} + 19 {x}^{3} - {x}^{2} + 19 x - 5} = \left(4 x - 1\right) \left(x - i\right) \left(x + i\right) \left(x + 5\right)$

So the rational zeros are $\frac{1}{4}$ and $- 5$. The other two zeros are $\pm i$.