How do you use the rational roots theorem to find all possible zeros of $f (x) = x^{3} - 10 x^{2} + 9 x - 24$ $f(x) = x^3 - 10x^2 + 9x - 24$ ?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = x^{3} - 10 x^{2} + 9 x - 24$ $f(x) = x^3 - 10x^2 + 9x - 24$ ?

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Answer 1 · 2016-08-13T19:24:53

Find $f (x)$ $f(x)$ has no rational zeros, but we can use Cardano's method to find Real zero:

$x_{1} = \frac{1}{3} (10 + \sqrt[3]{- 919 + 18 \sqrt{1406}} + \sqrt[3]{- 919 - 18 \sqrt{1406}})$ $x_1 = 1/3(10+root(3)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406)))$

and related Complex zeros.

Explanation:

$f (x) = x^{3} - 10 x^{2} + 9 x - 24$ $f(x) = x^3-10x^2+9x-24$

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 $- 24$ $-24$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term. That means that the only possible rational zeros are:

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$ $+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-24$

In addition, using Descartes' rule of signs we find that $f (x)$ $f(x)$ has $1$ $1$ or $3$ $3$ positive Real zeros and no negative Real zeros. So the only possible rational zeros are:

$1, 2, 3, 4, 6, 8, 12, 24$ $1, 2, 3, 4, 6, 8, 12, 24$

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

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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, $a=1$ , $b=-10$ , $c=9$ and $d=-24$ , so we find:

$Delta = 8100-2916-96000-15552+38880 = -67488$

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.

$0=27f(x)=27x^3-270x^2+243x-648$

$=(3x-10)^3-219(3x-10)-1838$

$=t^3-219t-1838$

where $t=(3x-10)$

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Cardano's method

We want to solve:

$t^3-219t-1838=0$

Let $t=u+v$ .

Then:

$u^3+v^3+3(uv-73)(u+v)-1838=0$

Add the constraint $v=73/u$ to eliminate the $(u+v)$ term and get:

$u^3+389017/u^3-1838=0$

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

$(u^3)^2-1838(u^3)+389017=0$

Use the quadratic formula to find:

$u^3=(1838+-sqrt((-1838)^2-4(1)(389017)))/(2*1)$

$=(-1838+-sqrt(3378244-1556068))/2$

$=(-1838+-sqrt(1822176))/2$

$=(-1838+-36sqrt(1406))/2$

$=-919+-18sqrt(1406)$

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:

$t_1=root(3)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406))$

and related Complex roots:

$t_2=omega root(3)(-919+18sqrt(1406))+omega^2 root(3)(-919-18sqrt(1406))$

$t_3=omega^2 root(3)(-919+18sqrt(1406))+omega root(3)(-919-18sqrt(1406))$

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

Now $x=1/3(10+t)$ . So the zeros of our original cubic are:

$x_1 = 1/3(10+root(3)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406)))$

$x_2 = 1/3(10+omega root(3)(-919+18sqrt(1406))+omega^2 root(3)(-919-18sqrt(1406)))$

$x_3 = 1/3(10+omega^2 root(3)(-919+18sqrt(1406))+omega root(3)(-919-18sqrt(1406)))$

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How do you use the rational roots theorem to find all possible zeros of $f (x) = x^{3} - 10 x^{2} + 9 x - 24$ $f(x) = x^3 - 10x^2 + 9x - 24$ ?

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Explanation:

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How do you use the rational roots theorem to find all possible zeros of f(x) = x^3 - 10x^2 + 9x - 24f(x)=x3−10x2+9x−24f(x) = x^3 - 10x^2 + 9x - 24?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = x^{3} - 10 x^{2} + 9 x - 24$ $f(x) = x^3 - 10x^2 + 9x - 24$ ?