# How do you find all the zeros of f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19?

Jul 31, 2016

Use Durand-Kerner method to find numerical approximations:

${x}_{1} \approx - 4.61495$

${x}_{2 , 3} \approx 1.34883 \pm 0.412784 i$

${x}_{4 , 5} \approx - 0.374682 \pm 0.74117 i$

#### Explanation:

$f \left(x\right) = - 3 {x}^{5} - 8 {x}^{4} + 25 {x}^{3} - 8 {x}^{2} + x - 19$

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $- 19$ and $q$ a divisor of the coefficient $- 3$ of the leading term.

So the only possible rational zeros are:

$\pm \frac{1}{3} , \pm 1 , \pm \frac{19}{3} , \pm 19$

Evaluating $f \left(x\right)$ for each of these values we find none works. So $f \left(x\right)$ has no rational zeros.

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In common with most quintics, the zeros of this $f \left(x\right)$ are not expressible in terms of $n$th roots. There are no algebraic solutions in terms of elementary functions, including trigonometric ones.

About the best we can do is use a numeric method such as Durand-Kerner to find approximations to the zeros:

Suppose the $5$ zeros are: $p , q , r , s , t$.

Choose initial approximations for these zeros as follows:

${p}_{0} = {\left(0.4 + 0.9 i\right)}^{0}$

${q}_{0} = {\left(0.4 + 0.9 i\right)}^{1}$

${r}_{0} = {\left(0.4 + 0.9 i\right)}^{2}$

${s}_{0} = {\left(0.4 + 0.9 i\right)}^{3}$

${t}_{0} = {\left(0.4 + 0.9 i\right)}^{4}$

Then iterate using the formulas:

${p}_{i + 1} = {p}_{i} - \frac{f \left({p}_{i}\right)}{\left({p}_{i} - {q}_{i}\right) \left({p}_{i} - {r}_{i}\right) \left({p}_{i} - {s}_{i}\right) \left({p}_{i} - {t}_{i}\right)}$

${q}_{i + 1} = {q}_{i} - \frac{f \left({q}_{i}\right)}{\left({q}_{i} - {p}_{i + 1}\right) \left({q}_{i} - {r}_{i}\right) \left({q}_{i} - {s}_{i}\right) \left({q}_{i} - {t}_{i}\right)}$

${r}_{i + 1} = {r}_{i} - \frac{f \left({r}_{i}\right)}{\left({r}_{i} - {p}_{i + 1}\right) \left({r}_{i} - {q}_{i + 1}\right) \left({r}_{i} - {s}_{i}\right) \left({r}_{i} - {t}_{i}\right)}$

${s}_{i + 1} = {s}_{i} - \frac{f \left({s}_{i}\right)}{\left({s}_{i} - {p}_{i + 1}\right) \left({s}_{i} - {q}_{i + 1}\right) \left({s}_{i} - {r}_{i + 1}\right) \left({s}_{i} - {t}_{i}\right)}$

t_(i+1) = t_i-(f(t_i))/((t_i-p_(i+1))(t_i-q_(i+1))(t_i-r_(i+1))(t_i-s_(i+1))

Keep iterating until the values are stable to the desired accuracy.

With this method, I found the following approximations for the zeros:

${x}_{1} \approx - 4.61495$

${x}_{2 , 3} \approx 1.34883 \pm 0.412784 i$

${x}_{4 , 5} \approx - 0.374682 \pm 0.74117 i$

Here's a sample C++ program that implements the algorithm for the example quintic: