# How do you find all the real and complex roots of #x^5 - 32 = 0#?

##### 3 Answers

You must mean the solutions to the equation. In the solution below I will show you how to solve this equation

#### Explanation:

Now, since

x =

x = 2

Since an odd exponent with a negative base always gives a negative number (example

**Exercises:**

- Solve for x in the following equations.

a)

b)

Hopefully you understand now, and best of luck in your continuation.

#### Explanation:

It is evident that

Since

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

We want to solve for

#x^4+2x^3+4x^2+8x+16=0#

I will admit that finding the complex roots is difficult and unintuitive (I resorted to using Google to find a way to factor this):

Notice that

Thus, we can say that

#x^4+2x^3+9x^2+8x+16-5x^2=0#

Which simplifies to be

#(x^2+x+4)^2-5x^2=0#

This can be factored as a difference of squares.

#(x^2+x+4+xsqrt5)(x^2+x+4-xsqrt5)=0#

These quadratic equations can be solved individually:

The first:

#x^2+(1+sqrt5)x+4=0#

#x=(-1-sqrt5+-sqrt((1+sqrt5)^2-16))/2#

#x=(-1-sqrt5+-sqrt(-10+2sqrt5))/2#

#x=(-1-sqrt5+-isqrt(10-2sqrt5))/2#

The second:

#x^2+(1-sqrt5)x+4=0#

#x=(sqrt5-1+-sqrt((1-sqrt5)^2-16))/2#

#x=(sqrt5-1+-sqrt(-10-2sqrt5))/2#

#x=(sqrt5-1+-isqrt(10+2sqrt5))/2#

#### Explanation:

The real solution is obviously 2.

Let's extend the equation to polar complexes:

MORE EXPLANATION OR ANOTHER WAY OF LOOKING AT IT:

An easy way to see this to imagine that the solution divided up around the circle of radius 2 in complex plane. Now divide the circle by 5 so the solutions will be:

1) Real

2) Complex

Find the real and imaginary part:

3) Complex

Find the real and imaginary part:

4) Complex

Find the real and imaginary part:

4) Complex

Find the real and imaginary part:

Hope this help...