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

Mar 16, 2016

The only rational zero is $x = 3$.

The other zeros are non-Real Complex roots.

Explanation:

The rational root theorem tells us that any rational zeros of $p \left(x\right)$ are expressible in the form $\frac{m}{n}$ for some integers $m$ and $n$ with $m$ a divisor of the constant term $- 6$ and $n$ a divisor of the coefficient $1$ of the leading term.

That is, the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 6$

Trying each in turn, we find:

$p \left(3\right) = 243 - 324 + 81 + 6 - 6 = 0$

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

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

By the rational root theorem, the only remaining possible rational zeros are:

$\pm 1$, $\pm 2$

None of these is a zero of ${x}^{4} - {x}^{3} + 2$, so there are no more rational zeros.

That's as far as we can go with the rational root theorem.

$\textcolor{w h i t e}{}$
Epilogue

It is possible to express the solutions of ${x}^{4} - {x}^{3} + 2 = 0$ algebraically, but it gets rather messy.

If you are interested, here's the beginning of the solution...

To solve ${x}^{4} - {x}^{3} + 2 = 0$, multiply by $2$ and substitute $t = \frac{1}{x}$ to get:

$0 = 4 {t}^{4} - 2 t + 2$

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

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

Equating coefficients and rearranging a little:

$b + c = {a}^{2} / 2$

$b - c = \frac{2}{a}$

$b c = 2$

Then:

${\left({a}^{2}\right)}^{2} / 4 = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = \frac{4}{{a}^{2}} + 8$

Hence:

${\left({a}^{2}\right)}^{3} - 32 \left({a}^{2}\right) - 16 = 0$

This cubic in ${a}^{2}$ has $3$ Real roots, one of which is positive, giving a pair of possible Real values for $a$.

Substitute ${a}^{2} = \frac{8}{3} \sqrt{6} \cos \theta$

Then:

$16 = {\left({a}^{2}\right)}^{3} - 32 \left({a}^{2}\right)$

$= {\left(\frac{8}{3} \sqrt{6} \cos \theta\right)}^{3} - 32 {\left(\frac{4}{3} \sqrt{6} \cos \theta\right)}^{2}$

$= \frac{256}{9} \sqrt{6} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right)$

$= \frac{256}{9} \sqrt{6} \cos \left(3 \theta\right)$

So:

$\cos \left(3 \theta\right) = \frac{3 \sqrt{6}}{32}$

So:

$\theta = \pm \frac{1}{3} \left(\arccos \left(\frac{3 \sqrt{6}}{32}\right) + 2 k \pi\right)$ for $k = 0 , 1 , 2$

So:

${a}^{2} = \frac{8}{3} \sqrt{6} \cos \left(\frac{1}{3} \left(\arccos \left(\frac{3 \sqrt{6}}{32}\right) + 2 k \pi\right)\right)$

for $k = 0 , 1 , 2$

This gives a positive value ($\approx 5.89$) for $k = 0$

So we can let

a = sqrt(8/3 sqrt(6) cos(1/3 arccos((3sqrt(6))/32))

Then $b = {a}^{2} / 2 + \frac{1}{a}$ and $c = {a}^{2} / 2 - \frac{1}{a}$

Hence we get quadratics to solve for $t$ and hence can find values for $x$.