How do you find all the zeros of $f (x) = x^{3} + 3 x^{2} - 5 x + 8$ $f(x)=x^3+3x^2-5x+8$ ?

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How do you find all the zeros of $f (x) = x^{3} + 3 x^{2} - 5 x + 8$ $f(x)=x^3+3x^2-5x+8$ ?

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Answer 1 · 2016-07-16T21:37:33

Use Cardano's method to find Real zero:

$x_{1} = \frac{1}{3} ⎛ ⎝ - 3 + \sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{2}} + \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{2}} ⎞ ⎠$ $x_1 = 1/3(-3+root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))$

and related Complex zeros.

Explanation:

$f (x) = x^{3} + 3 x^{2} - 5 x + 8$ $f(x) = x^3+3x^2-5x+8$

From the rational root theorem we can deduce that the only possible rational zeros are:

$\pm 1, \pm 2, \pm 4, \pm 8$ $+-1, +-2, +-4, +-8$

None of these work, so there are no rational zeros.

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, $a=1$ , $b=3$ , $c=-5$ and $d=8$ , so we find:

$Delta = 225+500-864-1728-2160=-4027$

Since $Delta < 0$ , this cubic has one Real zero and two non-Real Complex zeros. As a result, Cardano's method will work well.

We can simplify the problem by making the substitution $t=x+1$ :

$x^3+3x^2-5x+8$

$= (x^3+3x^2+3x+1)-8(x+1)+15$

$=(x+1)^3-8(x+1)+15$

$= t^3-8t+15$

Using Cardano's method, let $t=u+v$ , then we want to solve:

$u^3+v^3+(3uv-8)(u+v)+15 = 0$

Add the constraint $v = 8/(3u)$ to eliminate the term in $(u+v)$ and get:

$u^3+512/(27u^3)+15 = 0$

Multiply through by $27u^3$ and rearrange slightly to get a quadratic in $u^3$ :

$27(u^3)^2+405(u^3)+512 = 0$

Use the quadratic formula to find:

$u^3=(-405+-sqrt(405^2-4(27)(512)))/(2*27)$

$=(-405+-sqrt(164025-55296))/54$

$=(-405+-sqrt(108729))/54$

$=(-405+-3sqrt(12081))/54$

Hence Real zero:

$t_1 = root(3)((-405+3sqrt(12081))/54) + root(3)((-405-3sqrt(12081))/54)$

$= 1/3(root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))$

and related Complex zeros:

$t_2 = 1/3(omega root(3)((-405+3sqrt(12081))/2) + omega^2 root(3)((-405-3sqrt(12081))/2))$

$t_3 = 1/3(omega^2 root(3)((-405+3sqrt(12081))/2) + omega root(3)((-405-3sqrt(12081))/2))$

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

Then $x = t-1$ , hence zeros of $f(x)$ :

$x_1 = 1/3(-3+root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))$

$x_2 = 1/3(-3+omega root(3)((-405+3sqrt(12081))/2) + omega^2 root(3)((-405-3sqrt(12081))/2))$

$x_3 = 1/3(-3+omega^2 root(3)((-405+3sqrt(12081))/2) + omega root(3)((-405-3sqrt(12081))/2))$

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How do you find all the zeros of $f (x) = x^{3} + 3 x^{2} - 5 x + 8$ $f(x)=x^3+3x^2-5x+8$ ?

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How do you find all the zeros of f(x)=x^3+3x^2-5x+8f(x)=x3+3x2−5x+8f(x)=x^3+3x^2-5x+8?

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Explanation:

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How do you find all the zeros of $f (x) = x^{3} + 3 x^{2} - 5 x + 8$ $f(x)=x^3+3x^2-5x+8$ ?