How do you find the solutions to the equation x^4-6x^3+5x^2+30x-50=0?

Aug 14, 2016

This quartic has roots: $3 \pm i$ and $\pm \sqrt{5}$

Explanation:

$f \left(x\right) = {x}^{4} - 6 {x}^{3} + 5 {x}^{2} + 30 x - 50$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 50$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 5 , \pm 10 , \pm 25 , \pm 50$

None of these work. So $f \left(x\right)$ has no rational zeros.

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Tschirnhaus transformation

First make a linear substitution called a Tschirnhaus transformation in order to simplify the quartic:

$16 f \left(x\right) = 16 {x}^{4} - 96 {x}^{3} + 80 {x}^{2} + 480 x - 800$

$= {\left(2 x - 3\right)}^{4} - 34 {\left(2 x - 3\right)}^{2} + 144 \left(2 x - 3\right) - 143$

$= {t}^{4} - 34 {t}^{2} + 144 t - 143$

where $t = \left(2 x - 3\right)$

Since this quartic in $t$ is monic and has no ${t}^{3}$ term, it can be expressed as the product of two monic quadratics with opposite middle terms:

${t}^{4} - 34 {t}^{2} + 144 t - 143$

$= \left({t}^{2} - a x + b\right) \left({t}^{2} + a x + c\right)$

$= {t}^{4} + \left(b + c - {a}^{2}\right) {t}^{2} + \left(b - c\right) a t + b c$

Equating coefficients and rearranging a little, we get:

$\left\{\begin{matrix}b + c = {a}^{2} - 34 \\ b - c = \frac{144}{a} \\ b c = - 143\end{matrix}\right.$

So we find:

${\left({a}^{2} - 34\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = \frac{20736}{a} ^ 2 - 572$

Expanding, this becomes:

${\left({a}^{2}\right)}^{2} - 68 \left({a}^{2}\right) + 1156 = \frac{20736}{\left({a}^{2}\right)} - 572$

Multiply through by $\left({a}^{2}\right)$ and rearrange a little to get:

${\left({a}^{2}\right)}^{3} - 68 {\left({a}^{2}\right)}^{2} + 1728 \left({a}^{2}\right) - 20736 = 0$

By the rational roots theorem, the possible rational roots of this cubic are any factors of $20736 = {2}^{8} \cdot {3}^{4}$

That's rather a lot to try, but there are various methods to speed up the search and eventually find ${a}^{2} = 36$ is a zero.

Let $a = 6$

Then:

$b = \frac{1}{2} \left({a}^{2} - 34 + \frac{144}{a}\right) = \frac{1}{2} \left(36 - 34 + 24\right) = 13$

$c = \frac{1}{2} \left({a}^{2} - 34 - \frac{144}{a}\right) = \frac{1}{2} \left(36 - 34 - 24\right) = - 11$

Hmmm. We could have spotted that one fairly easily from the simultaneous equations by looking at possible factorings of $b c = - 143$.

Well however we get there we now have two quadratics to solve:

$0 = {t}^{2} - 6 t + 13 = {\left(t - 3\right)}^{2} + {2}^{2} = \left(t - 3 - 2 i\right) \left(t - 3 + 2 i\right)$

$0 = {t}^{2} + 6 t - 11 = {\left(t + 3\right)}^{2} - {\left(2 \sqrt{5}\right)}^{2} = \left(t + 3 - 2 \sqrt{5}\right) \left(t + 3 + 2 \sqrt{5}\right)$

Then $x = \frac{1}{2} \left(t + 3\right)$

So we get solutions to our original quartic:

$x = \frac{1}{2} \left(\left(3 \pm 2 i\right) + 3\right) = 3 \pm i$

$x = \frac{1}{2} \left(\left(- 3 \pm 2 \sqrt{5}\right) + 3\right) = \pm \sqrt{5}$