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

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I have been said that a #n^(th)# degree polynomal should have #n# zeroes. The only zero I could find in #p(x) = x^5=32# is #x= 2# .

I have been said that a

##### 3 Answers

Repeated zeroes.

#### Explanation:

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

#### Explanation:

First let's straighten out the terminology a bit:

The equation: *roots*, which are the five *zeros* of the function

One of those zeros is

We find:

#x^5-32 = (x-2)(x^4+2x^3+4x^2+8x+16)#

The remaining quartic has

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 = 2cos((2pi)/5)+2isin((2pi)/5)#

#x = 2cos((4pi)/5)+2isin((4pi)/5)#

#x = 2cos((6pi)/5)+2isin((6pi)/5)#

#x = 2cos((8pi)/5)+2isin((8pi)/5)#

**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

Put

The root

This equation has five complex roots. One is a real root (

(1) What is an imaginary number?

You cannot solve the equation

However, you can "imagine" the number that satisfies this.

The

A

If

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

Euler found that

If you put

Imaginary number is a bridge between trigonometric and exponential functions.

(2) What is the four imaginary numbers of

Acording to De Moivre's formula, the roots of

Substituting

The roots of