# Is it possible to factor y=x^3+2x^2-18x+4? If so, what are the factors?

Jun 2, 2016

$y = \left(x - 0.228698\right) \left(x + 5.44241\right) \left(x - 3.21371\right)$

#### Explanation:

According to Gauss, (Gauss, Carl Friedrich - 1799)
"...every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ roots". (wikipedia).

So the polynomial

${p}_{3} \left(x\right) = {x}^{3} + 2 {x}^{2} - 18 x + 4$ has exactly $3$ roots ${x}_{1} , {x}_{2} , {x}_{3}$

so the factors are

${p}_{3} \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

Use Cardano's rule to find the roots giving

$\left\{{x}_{1} = 0.228698 , {x}_{2} = - 5.44241 , {x}_{3} = 3.21371\right\}$

https://socratic.org/questions/how-do-you-find-all-the-real-and-complex-roots-of-x-3-x-2-x-2#270714

Jun 2, 2016

Use a trigonometric method to find:

${x}^{3} + 2 {x}^{2} - 18 x + 4 = \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

where

${x}_{0} \approx 3.21370759428$

${x}_{1} \approx - 5.44240577684$

${x}_{2} \approx 0.22869818256$

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} + 2 {x}^{2} - 18 x + 4$

The discriminant $\Delta$ of a cubic $a {x}^{3} + b {x}^{2} + c x + d$ 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 $a = 1$, $b = 2$, $c = - 18$, $d = 4$ and we find:

$\Delta = 1296 + 23328 - 128 - 432 - 2592 = 21472 > 0$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

In general Cardano's method will not be very helpful to find the zeros in such a case, since the resulting algebraic solution will include cube roots of Complex numbers.

First simplify to remove the square term:

$27 f \left(x\right) = 27 {x}^{3} + 54 {x}^{2} - 486 x + 108$

$= {\left(3 x + 2\right)}^{3} - 174 \left(3 x + 2\right) + 448$

Let $t = 3 x + 2$

We want to solve:

${t}^{3} - 174 t + 448 = 0$

Let $t = 2 \sqrt{58} \cos \theta$

The multiplier $2 \sqrt{58}$ is chosen to get $4 {\cos}^{3} \theta - 3 \cos \theta$ below...

Then:

$0 = {t}^{3} - 174 t + 448$

$= {\left(2 \sqrt{58} \cos \theta\right)}^{3} - 174 \left(2 \sqrt{58} \cos \theta\right) + 448$

$= 464 \sqrt{58} {\cos}^{3} \theta - 348 \sqrt{58} \cos \theta + 448$

$= 116 \sqrt{58} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) + 448$

$= 116 \sqrt{58} \cos 3 \theta + 448$

So $\cos 3 \theta = - \frac{448}{116 \sqrt{58}} = - \frac{112}{29 \sqrt{58}}$

So $3 \theta = \pm \arccos \left(- \frac{112}{29 \sqrt{58}}\right) + 2 k \pi$

So $\theta = \pm \frac{1}{3} \left(\arccos \left(- \frac{112}{29 \sqrt{58}}\right) + 2 k \pi\right)$

So $\cos \theta = \cos \left(\frac{1}{3} \left(\arccos \left(- \frac{112}{29 \sqrt{58}}\right) + 2 k \pi\right)\right)$

So $t = \sqrt{232} \cos \left(\frac{1}{3} \left(\arccos \left(- \frac{112}{29 \sqrt{58}}\right) + 2 k \pi\right)\right)$

So $x = \frac{1}{3} \left(\sqrt{232} \cos \left(\frac{1}{3} \left(\arccos \left(- \frac{112}{29 \sqrt{58}}\right) + 2 k \pi\right)\right) - 2\right)$

which takes distinct values for $k = 0 , 1 , 2.$

These are our $3$ Real zeros ${x}_{k}$

Then: ${x}^{3} + 2 {x}^{2} - 18 x + 4 = \left(x - {x}_{0}\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

${x}_{0} \approx 3.21370759428$

${x}_{1} \approx - 5.44240577684$

${x}_{2} \approx 0.22869818256$