# How do you use the rational roots theorem to find all possible zeros of y=2x^3+2x^2-23x-9?

Jul 7, 2016

Use a trigonometric method to find Real zeros:

${x}_{n} = \frac{1}{3} \left(\sqrt{142} \cos \left(\frac{1}{3} \arccos \left(\frac{8 \sqrt{142}}{2201}\right) + \frac{2 n \pi}{3}\right) - 1\right)$

for $n = 0 , 1 , 2$.

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + 2 {x}^{2} - 23 x - 9$

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Rational roots 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 $- 9$ and $q$ a divisor of the coefficient $2$ of the leading term.

So the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 3 , \pm \frac{9}{2} , \pm 9$

None of these works, so $f \left(x\right)$ has no rational 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, $a = 2$, $b = 2$, $c = - 23$ and $d = - 9$, so we find:

$\Delta = 2116 + 97336 + 288 - 8748 + 14904 = 105896$

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

As a result, pure algebraic methods like Cardano's method will yield results containing irreducible cube roots of Complex numbers. So choose to use a trigonometric solution instead.

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Tschirnhaus transformation

First, simplify the cubic using a linear substitution:

$108 f \left(x\right) = 216 {x}^{3} + 216 {x}^{2} - 2484 x - 972$

$= {\left(6 x + 2\right)}^{3} - 426 \left(6 x + 2\right) - 128$

$= {t}^{3} - 426 t - 128$

where $t = 6 x + 2$

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Trigonometric solution

We want to solve:

${t}^{3} - 426 t - 128 = 0$

Consider the substitution:

$t = k \cos \theta$

Then:

${t}^{3} - 426 t - 128$

$= {k}^{3} {\cos}^{3} \theta - 426 k \cos \theta - 128$

$= {k}^{3} / 4 \left(4 {\cos}^{3} \theta\right) - 142 k \left(3 \cos \theta\right) - 128$

Solve:

${k}^{3} / 4 = 142 k$

to get:

$k = \pm 2 \sqrt{142}$

Let $k = 2 \sqrt{142}$ to get:

$0 = {t}^{3} - 426 t - 128$

$= 284 \sqrt{142} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 128$

$= 284 \sqrt{142} \cos \left(3 \theta\right) - 128$

Hence:

$\cos \left(3 \theta\right) = \frac{128}{248 \sqrt{142}} = \frac{8 \sqrt{142}}{2201}$

Hence roots:

${t}_{n} = 2 \sqrt{142} \cos \left(\frac{1}{3} \arccos \left(\frac{8 \sqrt{142}}{2201}\right) + \frac{2 n \pi}{3}\right)$

giving distinct values for $n = 0 , 1 , 2$

Then $x = \frac{1}{6} \left(t - 2\right)$ so the zeros of the original cubic are:

${x}_{n} = \frac{1}{3} \left(\sqrt{142} \cos \left(\frac{1}{3} \arccos \left(\frac{8 \sqrt{142}}{2201}\right) + \frac{2 n \pi}{3}\right) - 1\right)$

for $n = 0 , 1 , 2$.