find all real roots:

x^4-6x^2+8=0

and the other polynomial is:
8x^3+1=0

for the first one, all real roots are 2, -2, and the square root of 2. i'm not sure of the second one. hope this helps.

For the first one :

You put X = x²
Then you have (variable change) :
X² - 6X + 8 = 0
And you use common tools to solve that (use delta). You find X = 2 or X = 4.
You have x² = X, so x = sqrt(X) or x = -sqrt(X).
Here, the solutions are sqrt(2), -sqrt(2), 2, -2.

For the second one, you simply have :
x^3 = -1/8
x = sqrt3(-1/8) = -1/2

To find the real roots of a polynomial equation, you can use a few techniques. Let's start by solving the first equation:

x^4 - 6x^2 + 8 = 0

You can notice that the equation is a quadratic equation in terms of x^2. Let's substitute y = x^2 to simplify the equation:

y^2 - 6y + 8 = 0

Now, it is a regular quadratic equation. To solve it, you can use either factoring, completing the square, or the quadratic formula. In this case, factoring is the most convenient method:

(y - 4)(y - 2) = 0

Now, set each factor equal to zero:

y - 4 = 0 --> y = 4
y - 2 = 0 --> y = 2

Since we had substituted y = x^2, we can now replace y back in the equation:

x^2 = 4 --> x = ±√4 --> x = ±2
x^2 = 2 --> x = ±√2

So the real roots of the first equation are x = -2, x = -√2, x = √2, and x = 2.

Now let's move on to the second equation:

8x^3 + 1 = 0

There is no straightforward method like factoring in this case. You can use numerical methods like Newton's method or the bisection method to find the approximate solutions. However, there doesn't seem to be a simple solution in this case.

Hence, the real roots of the given polynomial equation 8x^3 + 1 = 0 are not easily obtainable using elementary methods.