# How do you use the rational roots theorem to find all possible zeros of P(x) = 2x^4 + 10x^3 -3x^2 -8x +4?

Aug 20, 2016

We find there are no rational zeros, but it is possible to solve algebraically.

#### Explanation:

$P \left(x\right) = 2 {x}^{4} + 10 {x}^{3} - 3 {x}^{2} - 8 x + 4$

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 $4$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

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

Evaluating $P \left(x\right)$ for each of these values, we find that none is a zero. So $P \left(x\right)$ has no rational zeros.

In fact, it has two Real zeros and two non-Real Complex zeros. It is a little messy to solve algebraically, but it can be done...

Tschirnhaus transformation

First simplify the quartic using a linear substitution called a Tschirnhaus transformation...

$128 P \left(x\right) = 256 {x}^{4} + 1280 {x}^{3} - 384 {x}^{2} - 1024 x + 512$

$= {\left(4 x + 5\right)}^{4} - 174 {\left(4 x + 5\right)}^{2} + 984 \left(4 x + 5\right) - 683$

$= {t}^{4} - 174 {t}^{2} + 984 t - 683$

where $t = 4 x + 5$

Since this quartic in $t$ is monic with no ${t}^{3}$ term, it will factor as the product of two monic quadratics with opposite middle terms:

${t}^{4} - 174 {t}^{2} + 984 t - 683$

$= \left({t}^{2} - a t + b\right) \left({t}^{2} + a t + c\right)$

$= {t}^{4} + \left(b + c - {a}^{2}\right) {t}^{2} + \left(b - c\right) a t + b c$

Equating coefficients and rearranging slightly, we get:

$\left\{\begin{matrix}b + c = {a}^{2} - 174 \\ b - c = \frac{984}{a} \\ b c = - 683\end{matrix}\right.$

Hence:

${\left({a}^{2} - 174\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = {\left(\frac{984}{a}\right)}^{2} + 4 \left(- 683\right)$

Expanded:

${\left({a}^{2}\right)}^{2} - 348 \left({a}^{2}\right) + 30276 = \frac{968256}{\left({a}^{2}\right)} - 2732$

Multiply through by $\left({a}^{2}\right)$ and rearrange slightly to get:

${\left({a}^{2}\right)}^{3} - 348 {\left({a}^{2}\right)}^{2} + 33008 \left({a}^{2}\right) - 968256 = 0$

We can solve this cubic in ${a}^{2}$ using Cardano's method to find Real root:

${a}^{2} = \frac{4}{3} \left(87 + 2 \sqrt[3]{- 6885 + 60 \sqrt{1761}} + 2 \sqrt[3]{- 6885 - 60 \sqrt{1761}}\right) \approx 16.1274$

So we can let $a$ be the positive square root:

$a = \sqrt{\frac{4}{3} \left(87 + 2 \sqrt[3]{- 6885 + 60 \sqrt{1761}} + 2 \sqrt[3]{- 6885 - 60 \sqrt{1761}}\right)}$

Then:

$b = \frac{1}{2} \left({a}^{2} - 174 + \frac{984}{a}\right)$

$c = \frac{1}{2} \left({a}^{2} - 174 - \frac{984}{a}\right)$

Hence two quadratics to solve to find $t$ and hence $x$.

Messy, but not difficult.