# How do you find the zeros of f(z) = z^4+2z^3-5z^2-7z+20 ?

Nov 3, 2016

Use a numerical method to find approximations to the zeros:

${x}_{1 , 2} \approx 1.49358 \pm 0.773932 i$

${x}_{3 , 4} \approx - 2.49358 \pm 0.921833 i$

#### Explanation:

$f \left(z\right) = {z}^{4} + 2 {z}^{3} - 5 {z}^{2} - 7 z + 20$

By the rational roots theorem, any rational zeros of $f \left(z\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $20$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 5$, $\pm 10$, $\pm 20$

None of these works (in fact they all give positive values), so $f \left(z\right)$ has no rational zeros.

This is a fairly typical quartic with two complex conjugate pairs of roots. It is possible to solve algebraically, but gets very messy.

The way I would approach it algebraically would be:

• Use a Tschirnhaus transformation to derive a simplified quartic of the form: ${t}^{4} + p {t}^{2} + q t + r = 0$

• Consider a factorisation of this simplified quartic of the form $\left({t}^{2} - a t + b\right) \left({t}^{2} + a t + c\right)$

• Equate coefficients and use ${\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c$ to get a cubic in ${a}^{2}$

• If the resulting cubic has three Real irrational roots then use a $t = k \cos \theta$ substitution to find the solutions in trigonometric form. Otherwise use Cardano's method to find the one Real zero.

• Use the positive zero of the cubic as ${a}^{2}$ and the positive square root of that as $a$.

• Derive $b$ and $c$ to get two quadratics to solve.

${x}_{1 , 2} \approx 1.49358 \pm 0.773932 i$
${x}_{3 , 4} \approx - 2.49358 \pm 0.921833 i$