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

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

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Answer 1 · 2016-07-07T22:30:50

Use a trigonometric method to find Real zeros:

$x_{n} = \frac{1}{3} (\sqrt{142} cos (\frac{1}{3} arccos (\frac{8 \sqrt{142}}{2201}) + \frac{2 n π}{3}) - 1)$ $x_n = 1/3(sqrt(142) cos(1/3 arccos((8sqrt(142))/2201) + (2npi)/3) - 1)$

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

Explanation:

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

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Rational roots theorem

Any rational zeros of $f (x)$ $f(x)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integers $p$ $p$ , $q$ $q$ with $p$ $p$ a divisor of the constant term $- 9$ $-9$ and $q$ $q$ a divisor of the coefficient $2$ $2$ of the leading term.

So the only possible rational zeros are:

$\pm \frac{1}{2}, \pm 1, \pm \frac{3}{2}, \pm 3, \pm \frac{9}{2}, \pm 9$ $+-1/2, +-1, +-3/2, +-3, +-9/2, +-9$

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

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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, $a=2$ , $b=2$ , $c=-23$ 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:

$108 f(x) = 216x^3+216x^2-2484x-972$

$=(6x+2)^3-426(6x+2)-128$

$=t^3-426t-128$

where $t = 6x+2$

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Trigonometric solution

We want to solve:

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

Consider the substitution:

$t = k cos theta$

Then:

$t^3-426t-128$

$= k^3 cos^3 theta - 426 k cos theta - 128$

$= k^3/4 (4cos^3 theta) - 142k (3cos theta) - 128$

Solve:

$k^3/4 = 142k$

to get:

$k = +-2sqrt(142)$

Let $k = 2sqrt(142)$ to get:

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

$= 284 sqrt(142)(4 cos^3theta - 3cos theta)-128$

$= 284 sqrt(142)cos(3 theta)-128$

Hence:

$cos(3 theta) = 128/(248sqrt(142)) = (8sqrt(142))/2201$

Hence roots:

$t_n =2 sqrt(142) cos(1/3 arccos((8sqrt(142))/2201) + (2npi)/3)$

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

Then $x = 1/6(t-2)$ so the zeros of the original cubic are:

$x_n = 1/3(sqrt(142) cos(1/3 arccos((8sqrt(142))/2201) + (2npi)/3) - 1)$

for $n = 0, 1, 2$ .

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

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How do you use the rational roots theorem to find all possible zeros of y=2x^3+2x^2-23x-9y=2x3+2x2−23x−9y=2x^3+2x^2-23x-9?

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