How do you find all the zeros of $f (x) = x^{3} + x^{2} - 11 x - 30$ $f(x)=x^3+x^2-11x-30$ ?

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How do you find all the zeros of $f (x) = x^{3} + x^{2} - 11 x - 30$ $f(x)=x^3+x^2-11x-30$ ?

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Answer 1 · 2016-08-12T16:25:16

Use Cardano's method to find Real zero:

$x_{1} = \frac{1}{3} ⎛ ⎝ - 1 + \sqrt[3]{\frac{- 709 + 9 \sqrt{4265}}{2}} + \sqrt[3]{\frac{- 709 - 9 \sqrt{4265}}{2}} ⎞ ⎠$ $x_1 = 1/3(-1+root(3)((-709+9sqrt(4265))/2)+root(3)((-709-9sqrt(4265))/2))$

and related Complex zeros.

Explanation:

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$f (x) = x^{3} + x^{2} - 11 x - 30$ $f(x) = x^3+x^2-11x-30$

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Descriminant

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, $a=1$ , $b=1$ , $c=-11$ and $d=-30$ , so we find:

$Delta = 121+5324+120-24300+5940 = -12795$

Since $Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

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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.

$0=27f(x)=27x^3+27x^2-297x-810$

$=(3x+1)^3-102(3x+1)-709$

$=t^3-102t-709$

where $t=(3x+1)$

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Cardano's method

We want to solve:

$t^3-102t-709=0$

Let $t=u+v$ .

Then:

$u^3+v^3+3(uv-34)(u+v)-709=0$

Add the constraint $v=34/u$ to eliminate the $(u+v)$ term and get:

$u^3+39304/u^3-709=0$

Multiply through by $u^3$ and rearrange slightly to get:

$(u^3)^2-709(u^3)+39304=0$

Use the quadratic formula to find:

$u^3=(709+-sqrt((-709)^2-4(1)(39304)))/(2*1)$

$=(-709+-sqrt(502681-157216))/2$

$=(-709+-sqrt(345465))/2$

$=(-709+-9sqrt(4265))/2$

Since this is Real and the derivation is symmetric in $u$ and $v$ , we can use one of these roots for $u^3$ and the other for $v^3$ to find Real root:

$t_1=root(3)((-709+9sqrt(4265))/2)+root(3)((-709-9sqrt(4265))/2)$

and related Complex roots:

$t_2=omega root(3)((-709+9sqrt(4265))/2)+omega^2 root(3)((-709-9sqrt(4265))/2)$

$t_3=omega^2 root(3)((-709+9sqrt(4265))/2)+omega root(3)((-709-9sqrt(4265))/2)$

where $omega=-1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ .

Now $x=1/3(-1+t)$ . So the roots of our original cubic are:

$x_1 = 1/3(-1+root(3)((-709+9sqrt(4265))/2)+root(3)((-709-9sqrt(4265))/2))$

$x_2 = 1/3(-1+omega root(3)((-709+9sqrt(4265))/2)+omega^2 root(3)((-709-9sqrt(4265))/2))$

$x_3 = 1/3(-1+omega^2 root(3)((-709+9sqrt(4265))/2)+omega root(3)((-709-9sqrt(4265))/2))$

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How do you find all the zeros of $f (x) = x^{3} + x^{2} - 11 x - 30$ $f(x)=x^3+x^2-11x-30$ ?

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How do you find all the zeros of f(x)=x^3+x^2-11x-30f(x)=x3+x2−11x−30f(x)=x^3+x^2-11x-30?

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Explanation:

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How do you find all the zeros of $f (x) = x^{3} + x^{2} - 11 x - 30$ $f(x)=x^3+x^2-11x-30$ ?