SOLUTION: The question wants you to find the complex zeros and factor the polynomial in standard form: F(X) = x^3 - 1.

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: The question wants you to find the complex zeros and factor the polynomial in standard form: F(X) = x^3 - 1.      Log On


   

Question 381141: The question wants you to find the complex zeros and factor the polynomial in standard form: F(X) = x^3 - 1.
Answer by neatmath(302) About Me  (Show Source):
You can put this solution on YOUR website!

We can easily see that f%28x%29=x%5E3-1 is the difference of two cubes!

f%28x%29=x%5E3-1

f%28x%29=%28x-1%29%28x%5E2%2Bx%2B1%29

0=%28x-1%29%28x%5E2%2Bx%2B1%29 set f(x) equal to zero, so we have

%28x-1%29=0 or %28x%5E2%2Bx%2B1%29=0

x-1=0

x=1 or

%28x%5E2%2Bx%2B1%29=0

This will involve imaginary numbers because there are NO real numbers which satisfy this equation, but we can use the quadratic formula to find the complex solutions involving imaginary numbers:

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A1=-3.

The discriminant -3 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -3 is + or - sqrt%28+3%29+=+1.73205080756888.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B1+%29