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

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

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Answer 1 · 2016-05-18T18:07:54

Use Cardano's method to find Real zero:

$x_{1} = \frac{1}{3} ⎛ ⎝ 1 + \sqrt[3]{\frac{- 43 + 9 \sqrt{29}}{2}} + \sqrt[3]{\frac{- 43 - 9 \sqrt{29}}{2}} ⎞ ⎠$ $x_1 = 1/3(1+root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2))$

and Complex zeros.

Explanation:

Premultiply by $3^{3}$ $3^3$ to cut down on arithmetic involving fractions:

$0 = 3^{3} (x^{3} - x^{2} + 2 x + 1)$ $0 = 3^3 (x^3-x^2+2x+1)$

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

$= {(3 x - 1)}^{3} + 15 (3 x - 1) + 43$ $=(3x-1)^3+15(3x-1)+43$

Let $t = 3 x - 1$ $t = 3x-1$ and solve:

$t^{3} + 15 t + 43 = 0$ $t^3+15t+43 = 0$

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

$u^{3} + v^{3} + 3 (u v + 5) (u + v) + 43 = 0$ $u^3+v^3+3(uv+5)(u+v)+43 = 0$

Let $v = - \frac{5}{u}$ $v = -5/u$ to eliminate the term in $(u + v)$ $(u+v)$

$u^{3} - \frac{5^{3}}{u^{3}} + 43 = 0$ $u^3-5^3/u^3+43 = 0$

Multiply through by $u^{3}$ $u^3$ to get a quadratic in $u^{3}$ $u^3$ :

${(u^{3})}^{2} + 43 (u^{3}) - 125 = 0$ $(u^3)^2+43(u^3)-125 = 0$

Use the quadratic formula to find:

$u^{3} = \frac{- 43 \pm \sqrt{43^{2} + (4 \cdot 125)}}{2}$ $u^3 = (-43+-sqrt(43^2+(4*125)))/2$

$= \frac{- 43 \pm \sqrt{1849 + 500}}{2}$ $=(-43+-sqrt(1849+500))/2$

$= \frac{- 43 \pm \sqrt{2349}}{2}$ $=(-43+-sqrt(2349))/2$

$= \frac{- 43 \pm 9 \sqrt{29}}{2}$ $=(-43+-9sqrt(29))/2$

The derivation was symmetric in $u$ $u$ and $v$ $v$ , so we can use one of these roots for $u^{3}$ $u^3$ and the other for $v^{3}$ $v^3$ to find the Real root of our cubic equation in $t$ $t$ is:

$t_{1} = \sqrt[3]{\frac{- 43 + 9 \sqrt{29}}{2}} + \sqrt[3]{\frac{- 43 - 9 \sqrt{29}}{2}}$ $t_1 = root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2)$

and Complex roots:

$t_{2} = ω \sqrt[3]{\frac{- 43 + 9 \sqrt{29}}{2}} + ω^{2} \sqrt[3]{\frac{- 43 - 9 \sqrt{29}}{2}}$ $t_2 = omega root(3)((-43+9sqrt(29))/2)+omega^2 root(3)((-43-9sqrt(29))/2)$

$t_{3} = ω^{2} \sqrt[3]{\frac{- 43 + 9 \sqrt{29}}{2}} + ω \sqrt[3]{\frac{- 43 - 9 \sqrt{29}}{2}}$ $t_3 = omega^2 root(3)((-43+9sqrt(29))/2)+omega root(3)((-43-9sqrt(29))/2)$

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 $x = \frac{1}{3} (t + 1)$ $x= 1/3(t+1)$ , hence zeros of the original cubic in $x$ $x$ :

$x_{1} = \frac{1}{3} ⎛ ⎝ 1 + \sqrt[3]{\frac{- 43 + 9 \sqrt{29}}{2}} + \sqrt[3]{\frac{- 43 - 9 \sqrt{29}}{2}} ⎞ ⎠$ $x_1 = 1/3(1+root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2))$

$x_{2} = \frac{1}{3} ⎛ ⎝ 1 + ω \sqrt[3]{\frac{- 43 + 9 \sqrt{29}}{2}} + ω^{2} \sqrt[3]{\frac{- 43 - 9 \sqrt{29}}{2}} ⎞ ⎠$ $x_2 = 1/3(1+omega root(3)((-43+9sqrt(29))/2)+omega^2 root(3)((-43-9sqrt(29))/2))$

$x_{3} = \frac{1}{3} ⎛ ⎝ 1 + ω^{2} \sqrt[3]{\frac{- 43 + 9 \sqrt{29}}{2}} + ω \sqrt[3]{\frac{- 43 - 9 \sqrt{29}}{2}} ⎞ ⎠$ $x_3 = 1/3(1+omega^2 root(3)((-43+9sqrt(29))/2)+omega root(3)((-43-9sqrt(29))/2))$

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