Verified answer

if 2 + √3 is a root, then 2 - √3 is a root as well.

Find the quadratic equation that has these two roots:

x^2 - px + q = 0, where p is the sum of the roots and q is the product.

x^2 - 4x + 1 = 0

this means that x^2 - 4x + 1 is a factor of 6x^4 - 13x^3 -35x^2 - x +3.

use long division and you'll find that

6x^4 - 13x^3 -35x^2 - x +3 = (x^2 - 4x + 1)(6x^2 + 11x + 3) = (x^2 - 4x + 1)(2x + 3)(3x + 1)

we have

(x^2 - 4x + 1)(2x + 3)(3x + 1) = 0

we already found out the (x^2 - 4x + 1) part so

(2x + 3)(3x + 1) = 0

x = -3/2 or x = -1/3

so there are 4 roots:

x = 2 + √3

x = 2 - √3

x = -3/2

x = -1/3

2 equivalent roots potential there's a turning element on the double root. resolve for dy/dx=0. fixing the quadratic 3x^2-8x-3=0 supplies x=3 or x=-a million/3. Checking the cubic shows the double root to be x=3.