1. ## Question

I have a question about the equation x^3+1=0 According to the Corollary of the fundamental theorem The degree of the the polynomial determines the amount of roots you get at the end regardless of whether they're Real or Complex.

For example the highest variable exponent is 3. So there should be 3 answers for x. All I found is -1, there should be 2 more answers.

Could someone find the other roots for me and explain how to get to them. Thanks

2. Originally Posted by redlight42
I have a question about the equation x^3+1=0 According to the Corollary of the fundamental theorem The degree of the the polynomial determines the amount of roots you get at the end regardless of whether they're Real or Complex.

For example the highest variable exponent is 3. So there should be 3 answers for x. All I found is -1, there should be 2 more answers.

Could someone find the other roots for me and explain how to get to them. Thanks
Hi redlight42,

It's like this:

$x^3+1=0$

$(x+1)$ is a factor, and simple polynomial division will reveal that $(x^2-x+1$ is the other factor.

$(x+1)(x^2-x+1)=0$

You found that $x+1=0$ gives you $x=-1$.

Use the quadratic formula on $x^2-x+1=0$ to find the two imaginary solutions.

3. Originally Posted by redlight42
I have a question about the equation x^3+1=0 According to the Corollary of the fundamental theorem The degree of the the polynomial determines the amount of roots you get at the end regardless of whether they're Real or Complex.

For example the highest variable exponent is 3. So there should be 3 answers for x. All I found is -1, there should be 2 more answers.

Could someone find the other roots for me and explain how to get to them. Thanks
Use the sum of two cubes to get the two factors:

$x^3+1 = (x+1)(x^2-x+1) = 0$

Evidently $x=-1$ is a solution and the quadratic formula gives solutions of $x = \frac{1}{2} \pm \frac{i \sqrt5}{2}$