# What are all the possible rational zeros for f(x)=x^5+2x^4+14x^3+26x^2+40x+80 and how do you find all zeros?

Aug 26, 2016

Find there are no rational zeros. Can use a numerical method to find approximations.

#### Explanation:

$f \left(x\right) = {x}^{5} + 2 {x}^{4} + 14 {x}^{3} + 26 {x}^{2} + 40 x + 80$

By the rational roots 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 $80$ and $q$ a divisor of the coefficient $1$ of the leading term.

In addition note that all of the coefficients of $f \left(x\right)$ are positive, so it has no positive zeros.

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

$- 1 , - 2 , - 4 , - 5 , - 8 , - 10 , - 16 , - 20 , - 40 , - 80$

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

Typically for a quintic or polynomial of higher degree, the zeros of $f \left(x\right)$ are not expressible in terms of $n$th roots or elementary functions (including trigonometric functions).

About the best you can do is find numerical approximations using a method such as Durand-Kerner to find:

${x}_{1} \approx - 1.92974$

${x}_{2 , 3} \approx - 0.135287 \pm 3.0955 i$

${x}_{4 , 5} \approx 0.100156 \pm 2.07561 i$

For more details on using Durand-Kerner to find such zeros of a quintic, see https://socratic.org/s/axney67c

Here's an example C++ program that implements the Durand-Kerner algorithm for our current example...