# How do you find all the zeros of f(x)= -4x^8-9x^7+9x-4 with its multiplicities?

Apr 16, 2016

See explanation...

#### Explanation:

This seems a strangely difficult problem to be given, but we can at least reduce it from an octic to a quartic, using the semi-symmetry of the coefficients...

$f \left(x\right) = - 4 {x}^{8} - 9 {x}^{7} + 9 x - 4$

Then:

$f \left(i z\right) = - 4 {\left(i z\right)}^{8} - 9 {\left(i z\right)}^{7} + 9 \left(i z\right) - 4$

$= - 4 {z}^{8} + 9 i {z}^{7} + 9 i z - 4$

$= {z}^{4} \left(- 4 {z}^{4} + 9 i {z}^{3} + 9 i {z}^{- 3} - 4 {z}^{- 4}\right)$

Let $t = z + \frac{1}{z}$

$f \frac{i z}{z} ^ 4 = - 4 {t}^{4} + 9 i {t}^{3} + 16 {t}^{2} - 27 i t - 8$

This quartic is not very simple to solve, but it is at least possible to solve algebraically. Each zero ${t}_{1} , {t}_{2} , {t}_{3} , {t}_{4}$ of this quartic gives you a quadratic in $z$ to solve and hence find the zeros of the original octic polynomial.

For example, one of the zeros of this quartic is:

$t = \frac{9 i}{16} - \frac{1}{16} \sqrt{\frac{1}{3} \left(269 - \frac{1424}{3 \sqrt{29534277} - 16282} ^ \left(\frac{1}{3}\right) + 16 {\left(3 \sqrt{29534277} - 16282\right)}^{\frac{1}{3}}\right)} - \frac{1}{2} \sqrt{\frac{269}{96} + \frac{89}{12 {\left(3 \sqrt{29534277} - 16282\right)}^{\frac{1}{3}}} - \frac{1}{12} {\left(3 \sqrt{29534277} - 16282\right)}^{\frac{1}{3}} + \frac{1881 i}{32 \sqrt{\frac{1}{3} \left(269 - \frac{1424}{3 \sqrt{29534277} - 16282} ^ \left(\frac{1}{3}\right) + 16 {\left(3 \sqrt{29534277} - 16282\right)}^{\frac{1}{3}}\right)}}}$