# How do you find the zeros of f(x) = 7x^3 + 4x^2 - 3x + 3?

Feb 19, 2016

Use a substitution to get into the form ${t}^{3} + p t + q = 0$ then Cardano's method...

#### Explanation:

When a cubic has one irrational Real zero and two non-Real Complex zeros, I like to solve using Cardano's method.

First, I would like to have a cubic with integer coefficients and no square term. To get there from our $f \left(x\right)$, multiply by ${3}^{3} \cdot {7}^{2} = 1323$ and substitute $t = 21 x + 4$ as follows:

$1323 \left(7 {x}^{3} + 4 {x}^{2} - 3 x + 3\right)$

$= 9261 {x}^{3} + 5292 {x}^{2} - 3969 x + 3969$

$= \left(9261 {x}^{3} + 5292 {x}^{2} + 1008 x + 64\right) - \left(4977 x + 948\right) + 4853$

$= \left({21}^{3} {x}^{3} + \left(3 \cdot {21}^{2} \cdot 4\right) {x}^{2} + \left(3 \cdot 21 \cdot {4}^{2}\right) x + {4}^{3}\right) - \left(\left(237 \cdot 21\right) x + \left(237 \cdot 4\right)\right) + 4853$

$= {\left(21 x + 4\right)}^{3} - 237 \left(21 x + 4\right) + 4853$

$= {t}^{3} - 237 t + 4853$

So we want to solve ${t}^{3} - 237 t + 4853 = 0$

Next substitute $t = u + v$ to get:

$0 = {\left(u + v\right)}^{3} - 237 \left(u + v\right) + 4853$

$= {u}^{3} + {v}^{3} + \left(3 u v - 237\right) \left(u + v\right) + 4853$

$= {u}^{3} + {v}^{3} + 3 \left(u v - 79\right) \left(u + v\right) + 4853$

Add the constraint $v = \frac{79}{u}$ and multiply through by ${u}^{3}$ to get:

${\left({u}^{3}\right)}^{2} + 4853 \left({u}^{3}\right) + 493039 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{- 4853 \pm \sqrt{{4853}^{2} - 4 \cdot 1 \cdot 493039}}{2}$

$= \frac{- 4853 \pm \sqrt{21579453}}{2}$

$= \frac{- 4853 \pm 63 \sqrt{5437}}{2}$

Since the derivation was symmetric in $u$ and $v$, one of these roots can be taken as ${u}^{3}$ and the other ${v}^{3}$ to give us the Real root:

${t}_{1} = \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}$

and the Complex roots:

${t}_{2} = \omega \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + {\omega}^{2} \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + \omega \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}$

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

Then $x = \frac{t - 4}{21}$, so the zeros of $f \left(x\right)$ are:

${x}_{1} = \frac{1}{21} \left(- 4 + \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}\right)$

${x}_{2} = \frac{1}{21} \left(- 4 + \omega \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + {\omega}^{2} \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}\right)$

${x}_{3} = \frac{1}{21} \left(- 4 + {\omega}^{2} \sqrt[3]{\frac{- 4853 + 63 \sqrt{5437}}{2}} + \omega \sqrt[3]{\frac{- 4853 - 63 \sqrt{5437}}{2}}\right)$