# How does x^5=32, a 5th degree polynomial, have 5 zeroes?

## I have been said that a ${n}^{t h}$ degree polynomal should have $n$ zeroes. The only zero I could find in $p \left(x\right) = {x}^{5} = 32$ is $x = 2$.

Sep 27, 2017

Repeated zeroes.

#### Explanation:

You are correct that the only zero present is $x = 2$, however, that zero is repeated because it is the only one present for the 5th degree polynomial. Essentially, the polynomial has 5 zeroes, all of which are $x = 2$.

Sep 30, 2017

${x}^{5} = 32$ has $5$ distinct roots, $x = 2$ and four non-real complex roots.

#### Explanation:

First let's straighten out the terminology a bit:

The equation: ${x}^{5} = 32$ has $5$ roots, which are the five zeros of the function ${x}^{5} - 32$.

One of those zeros is $x = 2$, which means that there is a corresponding factor $\left(x - 2\right)$.

We find:

${x}^{5} - 32 = \left(x - 2\right) \left({x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 8 x + 16\right)$

The remaining quartic has $4$ zeros, which are all non-real complex.

The five zeros form the vertices of a regular pentagon in the complex plane:

graph{((x-2)^2+y^2-0.01)((x-2cos(2pi/5))^2+(y-2sin(2pi/5))^2-0.01)((x-2cos(4pi/5))^2+(y-2sin(4pi/5))^2-0.01)((x-2cos(6pi/5))^2+(y-2sin(6pi/5))^2-0.01)((x-2cos(8pi/5))^2+(y-2sin(8pi/5))^2-0.01) = 0 [-5, 5, -2.5, 2.5]}

We can write them as:

$x = 2$

$x = 2 \cos \left(\frac{2 \pi}{5}\right) + 2 i \sin \left(\frac{2 \pi}{5}\right)$

$x = 2 \cos \left(\frac{4 \pi}{5}\right) + 2 i \sin \left(\frac{4 \pi}{5}\right)$

$x = 2 \cos \left(\frac{6 \pi}{5}\right) + 2 i \sin \left(\frac{6 \pi}{5}\right)$

$x = 2 \cos \left(\frac{8 \pi}{5}\right) + 2 i \sin \left(\frac{8 \pi}{5}\right)$

Notes

It is possible to solve the quartic to get expressions for these zeros in terms of square roots instead of trigonometric functions. One way to start is to use the factorisation into quadratics described in: https://socratic.org/questions/how-do-you-factor-x-5-y-5

${x}^{5} - {y}^{5} = \left(x - y\right) \left({x}^{2} + \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) x y + {y}^{2}\right) \left({x}^{2} + \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) x y + {y}^{2}\right)$

Put $y = 2$ for the given problem.

Oct 1, 2017

The root $x = 2$ is not repeted five times. If so, the fomula should be${\left(x - 2\right)}^{5} = 0$.

This equation has five complex roots. One is a real root ($x = 2$) and the other four are imaginary roots.

$\textcolor{red}{\text{Imaginary}}$ root is a root that is not a real number, but a complex one.

(1) What is an imaginary number?
You cannot solve the equation $\textcolor{b l u e}{{x}^{2} = - 1}$ if you are in the "real" world.
However, you can "imagine" the number that satisfies this.

The $\textcolor{red}{\text{imaginary unit}}$ $i$ is defined as $\textcolor{g r e e n}{{i}^{2} = - 1}$.
A $\textcolor{red}{\text{complex number}}$ $z$ is a number that has a form $\textcolor{b r o w n}{z = a + b i}$ ($a , b =$real numbers).
If $b$ is not zero, $z$ is an imaginary number.

Imaginary number is indeed "imaginary" but very useful for expressing an alternating current, dealing with rotational movement, etc.

Euler found that ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$. ($e = 2.718$… is the Napier's constant.)
If you put $\theta = \pi$, you will see the amazing result. $\textcolor{p u r p \le}{{e}^{i \pi} = - 1}$.
Imaginary number is a bridge between trigonometric and exponential functions.

(2) What is the four imaginary numbers of ${x}^{5} = 32$
Acording to De Moivre's formula, the roots of ${z}^{n} = 1$ have the form
z=cos((2πk)/n) + i sin((2πk)/n) ($k$=integer, $i$=imaginary unit).

Substituting $n = 5$ and $k = 0 , 1 , 2 , 3 , 4$ to this formula leads to the five roots of ${z}^{5} = 1$.
$z = \cos 0 + i \sin 0 = 1$ (the only real root), cos((2π)/5)±isin((2π)/5) and cos((4π)/5)±isin((4π)/5).

The roots of ${x}^{5} = 32$ are obtained by doubling $z$ ($x = 2 z$).