# How do you use the rational root theorem to list all possible roots for x^3+5x^2-2x-15=0?

Nov 18, 2016

The "possible" rational roots are $\pm 1 , \pm 3 , \pm 5 , \pm 15$

The actual roots are:

${x}_{k} = \frac{1}{3} \left(2 \sqrt{31} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{65 \sqrt{31}}{1922}\right) + \frac{2 k \pi}{3}\right) - 5\right)$

$k = 0 , 1 , 2$

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} + 5 {x}^{2} - 2 x - 15$

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Rational root theorem

The rational root theorem will only tell you about possible rational roots - not irrational or Complex ones.

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 15$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1 , \pm 3 , \pm 5 , \pm 15$

However, note that all of these are odd, resulting in all of the terms of $f \left(x\right)$ being odd, except $- 2 x$, which will be even. So for any of these values we have:

$f \left(x\right) = \text{odd" + "odd" + "even" + "odd" = "odd} \ne 0$

So $f \left(x\right)$ has no rational zeros.

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Bonus

Let us see what else we can find out about $f \left(x\right)$'s zeros...

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Discriminant

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, $a = 1$, $b = 5$, $c = - 2$ and $d = - 15$, so we find:

$\Delta = 100 + 32 + 7500 - 6075 + 2700 = 4257$

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

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Descartes's Rule of Signs

The signs of the coefficients of $f \left(x\right)$ are in the pattern $+ + - -$. With one change of sign, by Descartes' rule of signs, we can deduce that $f \left(x\right)$ has exactly one positive Real zero. So the other two Real zeros are negative.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 27 f \left(x\right) = 27 {x}^{3} + 135 {x}^{2} - 54 x - 405$

$= {\left(3 x + 5\right)}^{3} - 93 \left(3 x + 5\right) - 65$

$= {t}^{3} - 93 t - 65$

where $t = \left(3 x + 5\right)$

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Trigonometric substitution

Look for a substitution of the form:

$t = k \cos \theta$

where $k$ is chosen such that the resulting cubic in $\cos \theta$ contains:

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

Let $k = 2 \sqrt{31}$

Then:

$0 = {t}^{3} - 93 t - 65$

$\textcolor{w h i t e}{0} = {k}^{3} {\cos}^{3} \theta - 93 k \cos \theta - 65$

$\textcolor{w h i t e}{0} = 2 \sqrt{31} \left(124 {\cos}^{3} \theta - 93 \cos \theta\right) - 65$

$\textcolor{w h i t e}{0} = 62 \sqrt{31} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 65$

$\textcolor{w h i t e}{0} = 62 \sqrt{31} \cos 3 \theta - 65$

Hence:

$\cos 3 \theta = \frac{65}{62 \sqrt{31}} = \frac{65 \sqrt{31}}{1922}$

So:

$3 \theta = \pm {\cos}^{- 1} \left(\frac{65 \sqrt{31}}{1922}\right) + 2 k \pi$

$t = 2 \sqrt{31} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{65 \sqrt{31}}{1922}\right) + \frac{2 k \pi}{3}\right)$

Then $x = \frac{1}{3} \left(t - 5\right)$ and hence the zeros of our original cubic are:

${x}_{k} = \frac{1}{3} \left(2 \sqrt{31} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{65 \sqrt{31}}{1922}\right) + \frac{2 k \pi}{3}\right) - 5\right)$

$k = 0 , 1 , 2$