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

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

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Answer 1 · 2016-06-24T22:48:34

Find Real zero:

$x_{1} = \frac{1}{9} (2 + \sqrt[3]{- 262 + 9 \sqrt{849}} + \sqrt[3]{- 262 - 9 \sqrt{849}})$ $x_1 = 1/9(2+root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849)))$

and related non-Real Complex zeros using Cardano's method.

Explanation:

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

Rational root theorem

By the rational root 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 $2$ $2$ and $q$ $q$ a divisor of the coefficient $3$ $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3}$ $+-1/3$ , $\pm \frac{2}{3}$ $+-2/3$ , $\pm 1$ $+-1$ , $\pm 2$ $+-2$

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

Discriminant

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, $a=3$ , $b=-2$ , $c=1$ and $d=2$ , so:

$Delta = 4-12+64-972-216=-1132$

Since $Delta < 0$ , this cubic has $1$ Real zero and $2$ non-Real Complex zeros.

Cardano's method

We can use Cardano's method to find the zeros, but first we need to simplify the form of the cubic by premultiplying by $3^5=243$ and making a linear substitution (called a Tschirnhaus transformation).

$243 f(x) = 243(3x^3-2x^2+x+2)$

$= 729x^3-486x^2+243x+486$

$= (9x-2)^3+15(9x-2)+524$

Let $t=9x-2$

We want to solve:

$t^3+15t+524 = 0$

Let $t = u+v$

Then:

$u^3+v^3+3(uv+5)(u+v)+524=0$

Add the constraint:

$v = -5/u$

to eliminate the term in $(u+v)$ and get:

$u^3-125/u^3+524 = 0$

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

$(u^3)^2+524(u^3)-125 = 0$

Use the quadratic formula to find roots:

$u^3=(-524+-sqrt(524^2-4(1)(-125)))/2$

$=(-524+-sqrt(274576+500))/2$

$=(-524+-sqrt(275076))/2$

$=(-524+-18sqrt(849))/2$

$=-262+-9sqrt(849)$

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

$t_1 = root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849))$

and corresponding Complex roots:

$t_2 = omega root(3)(-262+9sqrt(849))+omega^2 root(3)(-262-9sqrt(849))$

$t_3 = omega^2 root(3)(-262+9sqrt(849))+omega root(3)(-262-9sqrt(849))$

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

Hence zeros for the original cubic:

$x_1 = 1/9(2+root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849)))$

$x_2 = 1/9(2+omega root(3)(-262+9sqrt(849))+omega^2 root(3)(-262-9sqrt(849)))$

$x_3 = 1/9(2+omega^2 root(3)(-262+9sqrt(849))+omega root(3)(-262-9sqrt(849)))$

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

1 Answer

Explanation:

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