How do you use the rational roots theorem to find all possible zeros of $f (x) = 2 x^{3} + 7 x^{2} + 8 x - 8$ $f(x)=2x^3+7x^2+8x-8$ ?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = 2 x^{3} + 7 x^{2} + 8 x - 8$ $f(x)=2x^3+7x^2+8x-8$ ?

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Answer 1 · 2016-08-11T11:21:07

Find $f (x)$ $f(x)$ has no rational zeros.

Use Cardano's method to find Real zero:

$x_{1} = \frac{1}{6} (- 7 + \sqrt[3]{- 593 + 12 \sqrt{2442}} + \sqrt[3]{- 593 - 12 \sqrt{2442}})$ $x_1 = 1/6(-7+root(3)(-593+12sqrt(2442))+root(3)(-593-12sqrt(2442)))$

and related Complex zeros.

Explanation:

$f (x) = 2 x^{3} + 7 x^{2} + 8 x - 8$ $f(x) = 2x^3+7x^2+8x-8$

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 $- 8$ $-8$ and $q$ $q$ a divisor of the coefficient $2$ $2$ of the leading term. So the only possible rational zeros are:

$\pm \frac{1}{2}, \pm 1, \pm 2, \pm 4, \pm 8$ $+-1/2, +-1, +-2, +-4, +-8$

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

We can find the irrational zeros using Cardano's method.

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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, $a=2$ , $b=7$ , $c=8$ and $d=-8$ , so we find:

$Delta = 3136-4096+10976-6912-16128 = -13024$

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.

$0=108f(x)=216x^3+756x^2+864x-864$

$=(6x+7)^3-3(6x+7)-1186$

$=t^3-3t-1186$

where $t=(6x+7)$

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Cardano's method

We want to solve:

$t^3-3t-1186=0$

Let $t=u+v$ .

Then:

$u^3+v^3+3(uv-1)(u+v)-1186=0$

Add the constraint $v=1/u$ to eliminate the $(u+v)$ term and get:

$u^3+1/u^3-1186=0$

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

$(u^3)^2-1186(u^3)+1=0$

Use the quadratic formula to find:

$u^3=(1186+-sqrt((-1186)^2-4(1)(1)))/(2*1)$

$=(-1186+-sqrt(1406596-4))/2$

$=(-1186+-sqrt(1406592))/2$

$=(-1186+-24sqrt(2442))/2$

$=-593+-12sqrt(2442)$

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:

$t_1=root(3)(-593+12sqrt(2442))+root(3)((-593-12sqrt(2442))$

and related Complex roots:

$t_2=omega root(3)(-593+12sqrt(2442))+omega^2 root(3)(-593-12sqrt(2442))$

$t_3=omega^2 root(3)(-593+12sqrt(2442))+omega root(3)(-593-12sqrt(2442))$

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

Now $x=1/6(-7+t)$ . So the zeros of our original cubic are:

$x_1 = 1/6(-7+root(3)(-593+12sqrt(2442))+root(3)(-593-12sqrt(2442)))$

$x_2 = 1/6(-7+omega root(3)(-593+12sqrt(2442))+omega^2 root(3)(-593-12sqrt(2442)))$

$x_3 = 1/6(-7+omega^2 root(3)(-593+12sqrt(2442))+omega root(3)(-593-12sqrt(2442)))$

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How do you use the rational roots theorem to find all possible zeros of $f (x) = 2 x^{3} + 7 x^{2} + 8 x - 8$ $f(x)=2x^3+7x^2+8x-8$ ?

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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)=2x^3+7x^2+8x-8f(x)=2x3+7x2+8x−8f(x)=2x^3+7x^2+8x-8?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = 2 x^{3} + 7 x^{2} + 8 x - 8$ $f(x)=2x^3+7x^2+8x-8$ ?