# How do you find the zeros of f(x) = 2x^3+2x^2-2x-1 ?

Oct 27, 2016

Use a trigonometric substitution method to find zeros:

${x}_{n} = \frac{1}{3} \left(4 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{5}{32}\right) + \frac{2 n \pi}{3}\right) - 1\right) \text{ }$

for $n = 0 , 1 , 2$.

#### Explanation:

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

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Rational roots theorem

By the 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 $- 1$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

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

None of these work, 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, $a = 2$, $b = 2$, $c = - 2$ and $d = - 1$, so we find:

$\Delta = 16 + 64 + 32 - 108 + 144 = 148$

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 using a linear substitution known as a Tschirnhaus transformation.

$0 = 108 f \left(x\right) = 216 {x}^{3} + 216 {x}^{2} - 216 x - 108$

$= {\left(6 x + 2\right)}^{3} - 48 \left(6 x + 2\right) - 20$

$= {t}^{3} - 48 t - 20$

where $t = \left(6 x + 2\right)$

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Trigonometric substitution

Look for a substitution of the form $t = k \cos \theta$ with $k$ chosen so that the resulting cubic contains:

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

Let $t = k \cos \theta$

$0 = {t}^{3} - 48 t - 20$

$\textcolor{w h i t e}{0} = {k}^{3} {\cos}^{3} \theta - 48 k \cos \theta - 20$

$\textcolor{w h i t e}{0} = k \left({k}^{2} {\cos}^{3} \theta - 48 \cos \theta\right) - 20$

Let $k = 8$

$0 = 8 \left(64 {\cos}^{3} \theta - 48 \cos \theta\right) - 20$

$\textcolor{w h i t e}{0} = 128 \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 20$

$\textcolor{w h i t e}{0} = 128 \cos 3 \theta - 20$

So:

$\cos 3 \theta = \frac{20}{128} = \frac{5}{32}$

Hence:

$3 \theta = \pm {\cos}^{- 1} \left(\frac{5}{32}\right) + 2 n \pi$

So:

$\theta = \pm \frac{1}{3} {\cos}^{- 1} \left(\frac{5}{32}\right) + \frac{2 n \pi}{3}$

Note that $\cos \left(- \theta\right) = \cos \left(\theta\right)$, so we can discard the $\pm$ to get distinct solutions:

${t}_{n} = 8 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{5}{32}\right) + \frac{2 n \pi}{3}\right) \text{ }$ for $n = 0 , 1 , 2$

Then $x = \frac{1}{6} \left(t - 2\right)$

${x}_{n} = \frac{1}{3} \left(4 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{5}{32}\right) + \frac{2 n \pi}{3}\right) - 1\right) \text{ }$ for $n = 0 , 1 , 2$