Giải phương trình: 2x^4 + 5x3 + 6x^2 - 10x + 8 = 0

To solve this equation, we can use the Rational Root Theorem to find possible rational roots. The possible rational roots are factors of the constant term (8) divided by factors of the leading coefficient (2).

The factors of 8 are ±1, ±2, ±4, ±8
The factors of 2 are ±1, ±2

Therefore, the possible rational roots are ±1, ±2, ±4, ±8.

We can then use synthetic division or polynomial long division to test each possible root. By testing, we find that x = 1 is a root of the equation.

By dividing the polynomial by (x - 1), we get:

2$x^{4}$ + 5$x^{3}$ + 6$x^{2}$ - 10x + 8 = (x - 1)(2$x^{3}$ + 7$x^{2}$ + 13x - 8)

Now, we have a cubic polynomial left to solve. We can continue using the Rational Root Theorem to find the remaining roots. By testing, we find that x = -2 is a root of the cubic polynomial.

By dividing the cubic polynomial by (x + 2), we get:

2$x^{3}$ + 7$x^{2}$ + 13x - 8 = (x + 2)(2$x^{2}$ + 3x - 4)

Now, we have a quadratic equation left to solve. We can use the quadratic formula to find the remaining roots:

2$x^{2}$ + 3x - 4 = 0

The solutions to this quadratic equation are:

x = (-3 ± √(3^2 - 4*2*(-4))) / 4
x = (-3 ± √(9 + 32)) / 4
x = (-3 ± √41) / 4

Therefore, the roots of the original equation 2$x^{4}$ + 5$x^{3}$ + 6$x^{2}$ - 10x + 8 = 0 are x = 1, x = -2, and x = (-3 ± √41) / 4.