How do you use the rational roots theorem to find all possible zeros of  p(x)=-5x^4+x^3+2x^2-1?

Aug 17, 2016

Find that this quartic has no rational solutions.

Show how to solve it algebraically regardless...

Explanation:

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

By the rational roots theorem, any rational zeros of $p \left(x\right)$ are expressible in the form $\frac{p}{q}$ for some integers $p , q$ with $p$ a divisor of the constant term $- 1$ and $q$ a divisor of the coefficient $- 5$ of the leading term.

That means that the only possible rational zeros are:

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

None of these work, so $p \left(x\right)$ has no rational zeros.

In fact, this is a fairly typical quartic, which is very messy to solve algebraically.

Here's a sketch of how to start...

Tschirnhaus transformation

First transform the quartic into a monic one with no cube term, using a linear substitution:

$- 32000 p \left(x\right) = 160000 {x}^{4} - 32000 {x}^{3} - 64000 {x}^{2} + 32000$

$= {\left(20 x - 1\right)}^{4} - 166 {\left(20 x - 1\right)}^{2} - 328 \left(20 x - 1\right) + 31837$

$= {t}^{4} - 166 {t}^{2} - 328 t + 31837$

where $t = \left(20 x - 1\right)$

Since our simplified quartic is monic with no cube term, it will factor as a product of two monic quadratics with opposite middle coefficients:

${t}^{4} - 166 {t}^{2} - 328 t + 31837$

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

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

Equating coefficients and rearranging a little, we get:

$\left\{\begin{matrix}b + c = {a}^{2} - 166 \\ b - c = - \frac{328}{a} \\ b c = 31837\end{matrix}\right.$

Hence we find:

${\left({a}^{2} - 166\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = {\left(- \frac{328}{a}\right)}^{2} + 4 \left(31837\right)$

Expand the two ends to get:

${\left({a}^{2}\right)}^{2} - 332 \left({a}^{2}\right) + 27556 = \frac{107584}{\left({a}^{2}\right)} + 127348$

Subtracting the right hand side from the left and multiplying through by ${a}^{2}$, we get:

${\left({a}^{2}\right)}^{3} - 332 {\left({a}^{2}\right)}^{2} - 99792 \left({a}^{2}\right) - 107584 = 0$

Note that if we can solve this cubic in $\left({a}^{2}\right)$ then we can find $b$ and $c$ using:

$b = \frac{1}{2} \left({a}^{2} - 166 - \frac{328}{a}\right)$

$c = \frac{1}{2} \left({a}^{2} - 166 + \frac{328}{a}\right)$

and hence have two quadratics to solve.

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Descriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 1$, $b = - 332$, $c = - 99792$ and $d = - 107584$, so we find:

$\Delta = 1097659450331136 + 3975091880804352 - 15747873947648 - 312506560512 - 64158470627328 = 4992532480000000$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler with another Tschirnhaus transformation.

$0 = 27 \left({\left({a}^{2}\right)}^{3} - 332 {\left({a}^{2}\right)}^{2} - 99792 \left({a}^{2}\right) - 107584\right)$

$= 27 {\left({a}^{2}\right)}^{3} - 8964 {\left({a}^{2}\right)}^{2} - 2694384 \left({a}^{2}\right) - 2904768$

$= {\left(3 \left({a}^{2}\right) - 332\right)}^{3} - 1228800 \left(3 \left({a}^{2}\right) - 332\right) - 374272000$

$= {s}^{3} - 1228800 s - 374272000$

where $s = \left(3 \left({a}^{2}\right) - 332\right)$

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Next steps

Since we know that this cubic has only Real zeros, methods like Cardano's method will result in expressions involving irreducible cube roots of Complex numbers. My preference in these cases is to solve using a trigonometric substitution:

$s = k \cos \theta$

where $k$ is chosen to squeeze the cubic into a form containing:

$4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$

I will stop here for now, but I think you will agree that such quartics are messy to solve.