# Solve the following equation in the complex number system. Please show all of your work. x^5+20x^3-x^2-20=0

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### 1 Answer

You need to form two groups of terms such that:

`(x^5 + 20x^3) - (x^2 + 20) = 0`

Notice that in the group `x^5 + 20x^3` you may factor out `x^3 ` such that:

`x^3(x^2 + 20) - (x^2 + 20) = 0`

You need to factor out `(x^2 + 20)` such that:

`(x^2 + 20)(x^3 - 1) = 0`

Hence `(x^2 + 20) = 0 =gt x^2 = -20 =gt x_(1,2) = +-sqrt(-20)` `x_(1,2) = +-2isqrt5` (complex number theory `sqrt(-1) = ` i)

`x^3 - 1 = 0`

You need to substitute `(x-1)(x^2 + x + 1)` for `x^3 - 1` such that:

`x^3 - 1 = 0 =gt (x-1)(x^2 + x + 1) = 0`

`x -1 = 0 =gt x = 1`

You need to use quadratic formula to solve `x^2 + x + 1 = 0` such that:

`x_(3,4) = (-1+-sqrt(1-4))/2`

`x_3 = (-1+isqrt3)/2 ; x_4 = (-1-isqrt3)/2`

**Hence, evaluating the complex solutions to the equation yields `x_1 = +2isqrt5; x_2 = -2isqrt5;x_3 = (-1+isqrt3)/2;x_4 = (-1-isqrt3)/2` .**