How do you find all the real and complex roots of F(x) = -9x^4 + 12x^3+40x^2-60x+25?

May 26, 2016

${x}_{1} = \frac{2}{3} + \frac{1}{3} i$

${x}_{2} = \frac{2}{3} - \frac{1}{3} i$

${x}_{3} = \sqrt{5}$

${x}_{4} = - \sqrt{5}$

Explanation:

Given:

$f \left(x\right) = - 9 {x}^{4} + 12 {x}^{3} + 40 {x}^{2} - 60 x + 25$

By the rational root theorem, any rational zeros of $f \left(x\right)$ will be expressible in the form $\frac{p}{q}$ with integers $p , q$, with $p$ a divisor of the constant term $25$ and $q$ a divisor of the coefficient $- 9$ of the leading term.

That means that the only possible rational zeros of $f \left(x\right)$ are:

$\pm \frac{1}{9} , \pm \frac{1}{3} , \pm \frac{5}{9} , \pm 1 , \pm \frac{5}{3} , \pm \frac{25}{9} , \pm 5 , \pm \frac{25}{3} , \pm 25$

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

Take a deep breath and start trying a full blown quartic solution:

Multiply through by $- 9$ to make the sums easier:

$- 9 f \left(x\right) = 81 {x}^{4} - 108 {x}^{3} + 360 {x}^{2} - 540 x + 225$

$= {\left(3 x - 1\right)}^{4} - 46 {\left(3 x - 1\right)}^{2} + 92 \left(3 x - 1\right) - 88$

Let $t = 3 x - 1$

We want to solve:

$0 = {t}^{4} - 46 {t}^{2} + 92 t - 88$

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

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

Equating coefficients and rearranging a little:

$\left\{\begin{matrix}b + c = {a}^{2} - 46 \\ b - c = \frac{92}{a} \\ b c = - 88\end{matrix}\right.$

Then:

${\left({a}^{2} - 46\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = {\left(\frac{92}{a}\right)}^{2} - 352$

Expanding both ends:

${a}^{4} - 92 {a}^{2} + 2116 = \frac{8464}{a} ^ 2 - 352$

Multiplying through by ${a}^{2}$ and rearranging a little, this becomes:

${\left({a}^{2}\right)}^{3} - 92 {\left({a}^{2}\right)}^{2} + 2468 \left({a}^{2}\right) - 8464 = 0$

Let $y = {a}^{2}$

Let $g \left(y\right) = {y}^{3} - 92 {y}^{2} + 2468 y - 8464$

By the rational root theorem, any rational zeros of $g \left(y\right)$ are factors of $8464$.

Trying the first few we find:

$g \left(4\right) = 64 - 1472 + 9872 - 8464 = 0$

So one solution is $y = 4$, that is ${a}^{2} = 4$. So we can put $a = 2$.

$\left\{\begin{matrix}b + c = {a}^{2} - 46 = {2}^{2} - 46 = - 42 \\ b - c = \frac{92}{a} = \frac{92}{2} = 46 \\ b c = - 88\end{matrix}\right.$

Hence:

$b = 2$ and $c = - 44$

So we have:

${t}^{4} - 46 {t}^{2} + 92 t - 88 = \left({t}^{2} - 2 t + 2\right) \left({t}^{2} + 2 t - 44\right)$

which has zeros:

${t}_{1} = 1 + i$

${t}_{2} = 1 - i$

${t}_{3} = - 1 + 3 \sqrt{5}$

${t}_{4} = - 1 - 3 \sqrt{5}$

Then $x = \frac{t + 1}{3}$, hence zeros of the original quartic:

${x}_{1} = \frac{2}{3} + \frac{1}{3} i$

${x}_{2} = \frac{2}{3} - \frac{1}{3} i$

${x}_{3} = \sqrt{5}$

${x}_{4} = - \sqrt{5}$