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

Jul 16, 2016

Use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{3} \left(- 3 + \sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{2}} + \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{2}}\right)$

and related Complex zeros.

#### Explanation:

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

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

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

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

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, $a = 1$, $b = 3$, $c = - 5$ 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 $t = x + 1$:

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

$= \left({x}^{3} + 3 {x}^{2} + 3 x + 1\right) - 8 \left(x + 1\right) + 15$

$= {\left(x + 1\right)}^{3} - 8 \left(x + 1\right) + 15$

$= {t}^{3} - 8 t + 15$

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

${u}^{3} + {v}^{3} + \left(3 u v - 8\right) \left(u + v\right) + 15 = 0$

Add the constraint $v = \frac{8}{3 u}$ to eliminate the term in $\left(u + v\right)$ and get:

${u}^{3} + \frac{512}{27 {u}^{3}} + 15 = 0$

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

$27 {\left({u}^{3}\right)}^{2} + 405 \left({u}^{3}\right) + 512 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{- 405 \pm \sqrt{{405}^{2} - 4 \left(27\right) \left(512\right)}}{2 \cdot 27}$

$= \frac{- 405 \pm \sqrt{164025 - 55296}}{54}$

$= \frac{- 405 \pm \sqrt{108729}}{54}$

$= \frac{- 405 \pm 3 \sqrt{12081}}{54}$

Hence Real zero:

${t}_{1} = \sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{54}} + \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{54}}$

$= \frac{1}{3} \left(\sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{2}} + \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{2}}\right)$

and related Complex zeros:

${t}_{2} = \frac{1}{3} \left(\omega \sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{2}} + {\omega}^{2} \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{2}}\right)$

${t}_{3} = \frac{1}{3} \left({\omega}^{2} \sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{2}} + \omega \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{2}}\right)$

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

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

${x}_{1} = \frac{1}{3} \left(- 3 + \sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{2}} + \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{2}}\right)$

${x}_{2} = \frac{1}{3} \left(- 3 + \omega \sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{2}} + {\omega}^{2} \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left(- 3 + {\omega}^{2} \sqrt[3]{\frac{- 405 + 3 \sqrt{12081}}{2}} + \omega \sqrt[3]{\frac{- 405 - 3 \sqrt{12081}}{2}}\right)$