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

Jun 24, 2016

Find Real zero:

${x}_{1} = \frac{1}{9} \left(2 + \sqrt[3]{- 262 + 9 \sqrt{849}} + \sqrt[3]{- 262 - 9 \sqrt{849}}\right)$

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

#### Explanation:

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

Rational root theorem

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3}$, $\pm \frac{2}{3}$, $\pm 1$, $\pm 2$

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

Discriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

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).

$243 f \left(x\right) = 243 \left(3 {x}^{3} - 2 {x}^{2} + x + 2\right)$

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

$= {\left(9 x - 2\right)}^{3} + 15 \left(9 x - 2\right) + 524$

Let $t = 9 x - 2$

We want to solve:

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

Let $t = u + v$

Then:

${u}^{3} + {v}^{3} + 3 \left(u v + 5\right) \left(u + v\right) + 524 = 0$

$v = - \frac{5}{u}$

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

${u}^{3} - \frac{125}{u} ^ 3 + 524 = 0$

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

${\left({u}^{3}\right)}^{2} + 524 \left({u}^{3}\right) - 125 = 0$

Use the quadratic formula to find roots:

${u}^{3} = \frac{- 524 \pm \sqrt{{524}^{2} - 4 \left(1\right) \left(- 125\right)}}{2}$

$= \frac{- 524 \pm \sqrt{274576 + 500}}{2}$

$= \frac{- 524 \pm \sqrt{275076}}{2}$

$= \frac{- 524 \pm 18 \sqrt{849}}{2}$

$= - 262 \pm 9 \sqrt{849}$

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:

${t}_{1} = \sqrt[3]{- 262 + 9 \sqrt{849}} + \sqrt[3]{- 262 - 9 \sqrt{849}}$

and corresponding Complex roots:

${t}_{2} = \omega \sqrt[3]{- 262 + 9 \sqrt{849}} + {\omega}^{2} \sqrt[3]{- 262 - 9 \sqrt{849}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{- 262 + 9 \sqrt{849}} + \omega \sqrt[3]{- 262 - 9 \sqrt{849}}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.

Hence zeros for the original cubic:

${x}_{1} = \frac{1}{9} \left(2 + \sqrt[3]{- 262 + 9 \sqrt{849}} + \sqrt[3]{- 262 - 9 \sqrt{849}}\right)$

${x}_{2} = \frac{1}{9} \left(2 + \omega \sqrt[3]{- 262 + 9 \sqrt{849}} + {\omega}^{2} \sqrt[3]{- 262 - 9 \sqrt{849}}\right)$

${x}_{3} = \frac{1}{9} \left(2 + {\omega}^{2} \sqrt[3]{- 262 + 9 \sqrt{849}} + \omega \sqrt[3]{- 262 - 9 \sqrt{849}}\right)$