How do you find all the zeros of $x^{3} + 2 x^{2} + 5 x + 1$ $x^3 + 2x^2 + 5x +1$ ?

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How do you find all the zeros of $x^{3} + 2 x^{2} + 5 x + 1$ $x^3 + 2x^2 + 5x +1$ ?

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Answer 1 · 2016-03-12T09:42:58

Use Cardano's method...

Explanation:

$f (x) = x^{3} + 2 x^{2} + 5 x + 1$ $f(x) = x^3+2x^2+5x+1$

To cut down on the number of fractions we need to work with, multiply by $3^{3} = 27$ $3^3 = 27$ first...

$0 = 27 f (x)$ $0 = 27f(x)$

$= 27 x^{3} + 54 x^{2} + 135 x + 27$ $=27x^3+54x^2+135x+27$

$= {(3 x + 2)}^{3} + 33 (3 x + 2) - 47$ $=(3x+2)^3+33(3x+2)-47$

Substitute $t = 3 x + 2$ $t = 3x + 2$ ...

$= t^{3} + 33 t - 47$ $=t^3+33t-47$

Using Cardano's method, substitute $t = u + v$ $t = u+v$ ...

$= u^{3} + v^{3} + 3 (u v + 11) (u + v) - 47$ $=u^3+v^3+3(uv+11)(u+v)-47$

Add the constraint $v = - \frac{11}{u}$ $v = -11/u$ to eliminate the $(u + v)$ $(u+v)$ term ...

$= u^{3} - (\frac{11^{3}}{u^{3}}) - 47$ $=u^3-(11^3/u^3)-47$

$= u^{3} - \frac{1331}{u^{3}} - 47$ $=u^3-1331/u^3-47$

Multiply through by $u^{3}$ $u^3$ to get this quadratic in $u^{3}$ $u^3$ ...

${(u^{3})}^{2} - 47 (u^{3}) - 1331 = 0$ $(u^3)^2-47(u^3)-1331 = 0$

Solve using the quadratic formula to get:

$u^{3} = \frac{47 \pm \sqrt{47^{2} + 4 \cdot 1331}}{2}$ $u^3 = (47+-sqrt(47^2+4*1331))/2$

$= \frac{47 \pm \sqrt{7533}}{2}$ $= (47+-sqrt(7533))/2$

$= \frac{47 \pm 9 \sqrt{93}}{2}$ $= (47+-9sqrt(93))/2$

Since the derivation was symmetric in $u$ $u$ and $v$ $v$ , we can use one of these roots for $u^{3}$ $u^3$ and the other for $v^{3}$ $v^3$ to derive the Real root:

$t_{1} = \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$ $t_1 = root(3)((47+9sqrt(93))/2) + root(3)((47-9sqrt(93))/2)$

and the Complex roots:

$t_{2} = ω \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + ω^{2} \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$ $t_2 = omega root(3)((47+9sqrt(93))/2) + omega^2 root(3)((47-9sqrt(93))/2)$

$t_{3} = ω^{2} \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + ω \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$ $t_3 = omega^2 root(3)((47+9sqrt(93))/2) + omega root(3)((47-9sqrt(93))/2)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ $1$

Then $x = \frac{t - 2}{3}$ $x = (t-2) / 3$ hence zeros of the original function:

$x_{1} = \frac{1}{3} ⎛ ⎝ - 2 + \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}} ⎞ ⎠$ $x_1 = 1/3(-2+root(3)((47+9sqrt(93))/2) + root(3)((47-9sqrt(93))/2) )$

$x_{2} = \frac{1}{3} ⎛ ⎝ - 2 + ω \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + ω^{2} \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}} ⎞ ⎠$ $x_2 = 1/3(-2 + omega root(3)((47+9sqrt(93))/2) + omega^2 root(3)((47-9sqrt(93))/2))$

$x_{3} = \frac{1}{3} ⎛ ⎝ - 2 + ω^{2} \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + ω \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}} ⎞ ⎠$ $x_3 = 1/3(-2 + omega^2 root(3)((47+9sqrt(93))/2) + omega root(3)((47-9sqrt(93))/2))$

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