What are all the zeroes of f(x) = 2x^3 + 3x^2 + 4x - 10?

Jan 11, 2016

$1.060419 \mathmr{and} - 1.280210 \pm i \cdot 1.753940$

Explanation:

Rewriting the function:
$f \left(x\right) = 2 \cdot \left({x}^{3} + 1.5 {x}^{2} + 2 x - 5\right)$

The easy way (trying as root $\pm 1 , \pm 5 \mathmr{and} \pm \frac{1}{5}$) doesn't help us to find the zeros.

Then we can go the hard way as described in
https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots

So for the general cubic equation

$a {x}^{3} + b {x}^{2} + c x + d = 0$

${x}_{k} = - \left(\frac{1}{3 a}\right) \left(b + {u}_{k} C + {\Delta}_{0} / \left({u}_{k} \cdot C\right)\right)$, $k = 1 , 2 , 3$

where
${u}_{1} = 1$, ${u}_{2} = \frac{- 1 + i \cdot \sqrt{3}}{2}$, ${u}_{3} = \frac{- 1 - i \cdot \sqrt{3}}{2}$

$C = \sqrt{\frac{{\Delta}_{1} + \sqrt{{\Delta}_{1}^{2} - 4 \cdot {\Delta}_{0}^{3}}}{2}}$ with
${\Delta}_{0} = {b}^{2} - 3 a c$
${\Delta}_{1} = 2 {b}^{3} - 9 a b c + 27 {a}^{2} d$

In case:
${\Delta}_{0} = {1.5}^{2} - 3 \cdot 1 \cdot 2 = 2.25 - 6 = - 3.75$
${\Delta}_{1} = 2 \cdot 3.375 - 9 \cdot 1 \cdot 1.5 \cdot 2 + 27 \cdot 1 \cdot \left(- 5\right) = 6.75 - 27 - 135 = - 155.25$

$C = \sqrt{\frac{- 155.25 + \sqrt{{\left(- 155.25\right)}^{2} - 4 \cdot {\left(- 3.75\right)}^{3}}}{2}}$
$C = \sqrt{\frac{- 155.25 + \sqrt{24313.5}}{2}} = \sqrt{.338934} = .697223$

For $k = 1 \mathmr{and} {u}_{1} = 1$
${x}_{1} = \left(- \frac{1}{3}\right) \left(1.5 + .697223 + \frac{- 3.75}{.697223}\right) = \left(- \frac{1}{3}\right) \left(2.197223 - 5.378480\right)$ => ${x}_{1} = 1.060419$

We could continue with this formula, but dividing $\left({x}^{3} + 1.5 {x}^{2} + 2 x - 5\right)$ by $\left(x - 1.060419\right)$ we get $\left({x}^{2} + 2.560419 x + 4.715117\right)$, whose roots we can obtain in this way:
$\left({x}^{2} + 2.560419 x + 4.715117\right) = 0$
$\Delta = 6.555745 - 18.860468 = - 12.304723$
$x = \frac{- 2.560 .419 \pm i \cdot 3.507809}{2}$ => ${x}_{2} = - 1.280210 + i \cdot 1.753904$, ${x}_{3} = - 1.280210 - i \cdot 1.753904$

Jan 16, 2016

${x}_{1} = \frac{1}{6} \left(- 3 + \sqrt{621 + 6 \sqrt{10806}} + \sqrt{621 - 6 \sqrt{10806}}\right)$

${x}_{2} = \frac{1}{6} \left(- 3 + \omega \sqrt{621 + 6 \sqrt{10806}} + {\omega}^{2} \sqrt{621 - 6 \sqrt{10806}}\right)$

${x}_{3} = \frac{1}{6} \left(- 3 + {\omega}^{2} \sqrt{621 + 6 \sqrt{10806}} + \omega \sqrt{621 - 6 \sqrt{10806}}\right)$

Explanation:

When a cubic has one Real and $2$ non-Real Complex zeros, I like to use Cardano's method...

Given $f \left(x\right) = 2 {x}^{3} + 3 {x}^{2} + 4 x - 10$

Let ${t}_{1} = 2 x$

Let ${t}_{2} = {t}_{1} + 1$

Then:

$4 f \left(x\right) = 8 {x}^{3} + 12 {x}^{2} + 16 x - 40$

$= {\left(2 x\right)}^{3} + 3 {\left(2 x\right)}^{2} + 8 \left(2 x\right) - 40$

$= {t}_{1}^{3} + 3 {t}_{1}^{2} + 8 {t}_{1} - 40$

$= \left({t}_{1}^{3} + 3 {t}_{1}^{2} + 3 {t}_{1} + 1\right) + \left(5 {t}_{1} + 5\right) - 46$

$= {\left({t}_{1} + 1\right)}^{3} + 5 \left({t}_{1} + 1\right) - 46$

$= {t}_{2}^{3} + 5 {t}_{2} - 46$

Next let ${t}_{2} = u + v$

$0 = {\left(u + v\right)}^{3} + 5 \left(u + v\right) - 46$

$= {u}^{3} + {v}^{3} + \left(3 u v + 5\right) \left(u + v\right) - 46$

To eliminate the term in $\left(u + v\right)$ add the constraint $v = - \frac{5}{3 u}$

$= {u}^{3} - {\left(\frac{5}{3 u}\right)}^{3} - 46$

$= {u}^{3} - \frac{125}{27 {u}^{3}} - 46$

Multiply through by $27 {u}^{3}$ and rearrange to obtain this quadratic in ${u}^{3}$:

$27 {\left({u}^{3}\right)}^{2} - 1242 \left({u}^{3}\right) - 125 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{1242 \pm \sqrt{{1242}^{2} + \left(4 \cdot 27 \cdot 125\right)}}{54}$

$= \frac{1242 \pm \sqrt{1556064}}{54}$

$= \frac{1242 \pm 12 \sqrt{10806}}{54}$

$= \frac{621 \pm 6 \sqrt{10806}}{27}$

So the Real value of $u$ is given by:

$u = \sqrt{\frac{621 \pm 6 \sqrt{10806}}{27}} = \frac{1}{3} \sqrt{621 \pm 6 \sqrt{10806}}$

Then since the derivation was symmetric in $u$ and $v$ we can deduce that one of these values is $u$ and the other $v$.

${t}_{2} = u + v$

$= \frac{1}{3} \sqrt{621 + 6 \sqrt{10806}} + \frac{1}{3} \sqrt{621 - 6 \sqrt{10806}}$

Hence the Real root of the original cubic is:

${x}_{1} = \frac{{t}_{2} - 1}{2}$

$= \frac{1}{6} \left(- 3 + \sqrt{621 + 6 \sqrt{10806}} + \sqrt{621 - 6 \sqrt{10806}}\right)$

The Complex roots come from replacing the Real value of $u$ with $\omega u$ or ${\omega}^{2} u$, where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive cube root of $1$ ...

${x}_{2} = \frac{1}{6} \left(- 3 + \omega \sqrt{621 + 6 \sqrt{10806}} + {\omega}^{2} \sqrt{621 - 6 \sqrt{10806}}\right)$

${x}_{3} = \frac{1}{6} \left(- 3 + {\omega}^{2} \sqrt{621 + 6 \sqrt{10806}} + \omega \sqrt{621 - 6 \sqrt{10806}}\right)$