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?

Question 187610This question is from textbook saxon algebra 2
: 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? This question is from textbook saxon algebra 2

Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): 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 (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

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 - .

The solution is , or
Here's your graph:

Answer by stanbon(75887) (Show Source): 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.
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