# How do you list all possible roots and find all factors of x^4+2x^3-8x^2+16x-23?

Dec 21, 2017

Here is a solution for ${x}^{4} - 2 {x}^{3} + 8 {x}^{2} + 16 x - 23 = 0$

(${x}^{4} + 2 {x}^{3} - 8 {x}^{2} + 16 x - 23 = 0$ is much more complicated)

#### Explanation:

Assuming typographical errors in the problem, let us solve:

${x}^{4} - 2 {x}^{3} + 8 {x}^{2} + 16 x - 23 = 0$

By the rational roots theorem, any rational zeros of this quartic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 23$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 23$

Note that $x = 1$ is a solution, since the sum of the coefficients is $0$, i.e. $1 - 2 + 8 + 16 - 23 = 0$

Then:

${x}^{4} - 2 {x}^{3} + 8 {x}^{2} + 16 x - 23 = \left(x - 1\right) \left({x}^{3} - {x}^{2} + 7 x + 23\right)$

To find the zeros for the remaining cubic, proceed as follows:

$f \left(x\right) = {x}^{3} - {x}^{2} + 7 x + 23$

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, $a = 1$, $b = - 1$, $c = 7$ and $d = 23$, so we find:

$\Delta = 49 - 1372 + 92 - 14283 - 2898 = - 18412$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 27 f \left(x\right) = 27 {x}^{3} - 27 {x}^{2} + 189 x + 621$

$= {\left(3 x - 1\right)}^{3} + 60 \left(3 x - 1\right) + 682$

$= {t}^{3} + 60 t + 682$

where $t = \left(3 x - 1\right)$

Cardano's method

We want to solve:

${t}^{3} + 60 t + 682 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v + 20\right) \left(u + v\right) + 682 = 0$

Add the constraint $v = - \frac{20}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} - \frac{8000}{u} ^ 3 + 682 = 0$

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

${\left({u}^{3}\right)}^{2} + 682 \left({u}^{3}\right) - 8000 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{- 682 \pm \sqrt{{\left(682\right)}^{2} - 4 \left(1\right) \left(- 8000\right)}}{2 \cdot 1}$

$= \frac{682 \pm \sqrt{465124 + 32000}}{2}$

$= \frac{682 \pm \sqrt{497124}}{2}$

$= 341 \pm 3 \sqrt{13809}$

Since this is Real and the derivation is symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to find Real root:

${t}_{1} = \sqrt[3]{341 + 3 \sqrt{13809}} + \sqrt[3]{341 - 3 \sqrt{13809}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{341 + 3 \sqrt{13809}} + {\omega}^{2} \sqrt[3]{341 - 3 \sqrt{13809}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{341 + 3 \sqrt{13809}} + \omega \sqrt[3]{341 - 3 \sqrt{13809}}$

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

Now $x = \frac{1}{3} \left(1 + t\right)$. So the roots of our original cubic are:

${x}_{1} = \frac{1}{3} \left(1 + \sqrt[3]{341 + 3 \sqrt{13809}} + \sqrt[3]{341 - 3 \sqrt{13809}}\right)$

${x}_{2} = \frac{1}{3} \left(1 + \omega \sqrt[3]{341 + 3 \sqrt{13809}} + {\omega}^{2} \sqrt[3]{341 - 3 \sqrt{13809}}\right)$

${x}_{3} = \frac{1}{3} \left(1 + {\omega}^{2} \sqrt[3]{341 + 3 \sqrt{13809}} + \omega \sqrt[3]{341 - 3 \sqrt{13809}}\right)$

Then the complete factorisation of our original quartic takes the form:

${x}^{4} - 2 {x}^{3} + 8 {x}^{2} + 16 x - 23 = \left(x - 1\right) \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$