SOLUTION: Find the roots of the polynomial function. y=x^3+8 y=x^3+8=(x+2)(x^2-2x+4) (-2)^2-4(1)(4)=12 one real root= -2 Could someone please check this for me?

You can put this solution on YOUR website!
Find the roots of the polynomial function.
y=x^3+8
y=x^3+8=(x+2)(x^2-2x+4)
(-2)^2-4(1)(4)=12
one real root= -2
Could someone please check this for me?
-----------------
y = (-2)^3 + 8
y = -8 + 8 = 0
It fits.
You don't need the complex roots? It says "Find the roots..."

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation ax%5E2%2Bbx%2Bc=0

(in our case 1x%5E2%2B-2x%2B4+=+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

.

The discriminant -12 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 -12 is + or - sqrt%28+12%29+=+3.46410161513775

.

The solution is x%5B12%5D+=+%28--2%2B-i%2Asqrt%28+-12+%29%29%2F2%5C1+=++%28--2%2B-i%2A3.46410161513775%29%2F2%5C1+

, or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B4+%29

You can put this solution on YOUR website!
Find the roots of the polynomial function.
y=x^3+8
y=x^3+8=(x+2)(x^2-2x+4)
-------------
x+2=0 or x^2-2x+4 = 0
x = -2 or Use the quadratic Formula:
---------------------------
x = [2 +- sqrt(4-4*1*4)]/2
x = [2 +- sqrt(-12)]/2
x = [2 +- 2isqrt(3)]/2
x = [1 +- isqrt(3)
=================================
Cheers,
Stan H.