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

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

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Answer 1 · 2016-10-27T20:18:09

Use a trigonometric substitution method to find zeros:

$x_{n} = \frac{1}{3} (4 cos (\frac{1}{3} {cos}^{- 1} (\frac{5}{32}) + \frac{2 n π}{3}) - 1)$ $x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "$

for $n = 0, 1, 2$ $n = 0, 1, 2$ .

Explanation:

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

$color(white)()$
Rational roots theorem

By the rational roots theorem, any rational zeros of $f (x)$ $f(x)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $- 1$ $-1$ and $q$ $q$ a divisor of the coefficient $2$ $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2}$ $+-1/2$ , $\pm 1$ $+-1$

None of these work, so $f (x)$ $f(x)$ has no rational zeros.

$color(white)()$
Discriminant

The discriminant $Delta$ of a cubic polynomial in the form $ax^3+bx^2+cx+d$ is given by the formula:

$Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$

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.

$0=108f(x)=216x^3+216x^2-216x-108$

$=(6x+2)^3-48(6x+2)-20$

$=t^3-48t-20$

where $t=(6x+2)$

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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:

$4cos^3 theta - 3 cos theta = cos 3 theta$

Let $t = k cos theta$

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

$color(white)(0) = k^3 cos^3 theta - 48k cos theta - 20$

$color(white)(0) = k (k^2 cos^3 theta - 48 cos theta) - 20$

Let $k=8$

$0 = 8 (64 cos^3 theta - 48 cos theta) - 20$

$color(white)(0) = 128 (4 cos^3 theta - 3 cos theta) - 20$

$color(white)(0) = 128 cos 3 theta - 20$

So:

$cos 3 theta = 20/128 = 5/32$

Hence:

$3 theta = +-cos^(-1)(5/32) + 2npi$

So:

$theta = +-1/3 cos^(-1)(5/32) + (2npi)/3$

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

$t_n = 8 cos (1/3 cos^(-1)(5/32) + (2npi)/3)" "$ for $n = 0, 1, 2$

Then $x = 1/6(t-2)$

$x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "$ for $n = 0, 1, 2$

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

1 Answer

Explanation:

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How do you find the zeros of f(x) = 2x^3+2x^2-2x-1f(x)=2x3+2x2−2x−1f(x) = 2x^3+2x^2-2x-1 ?

1 Answer

Explanation:

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