Find the roots of the equation x^3 - 2x^2 - 3x + 4 = 0.

The roots of x^3 - 2x^2 - 3x + 4 = 0 are to be found.

To find the roots of this equation, we can use the Rational Root Theorem to identify possible rational roots. The Rational Root Theorem states that if a polynomial has integer coefficients, then any rational root of the polynomial must have a numerator that divides the constant term and a denominator that divides the leading coefficient.

In this case, the constant term is 4 and the leading coefficient is 1, so the possible rational roots are ±1, ±2, ±4. We can test these roots by substituting them into the equation and checking if the result is zero.

Substituting x = 1, we get:
1^3 - 2(1)^2 - 3(1) + 4 = 0
So x = 1 is a root.

Substituting x = -1, we get:
(-1)^3 - 2(-1)^2 - 3(-1) + 4 = 0
So x = -1 is a root.

Substituting x = 4, we get:
4^3 - 2(4)^2 - 3(4) + 4 = 0
So x = 4 is not a root.

Substituting x = -4, we get:
(-4)^3 - 2(-4)^2 - 3(-4) + 4 = 0
So x = -4 is not a root.

Substituting x = 2, we get:
2^3 - 2(2)^2 - 3(2) + 4 = 0
So x = 2 is not a root.

Substituting x = -2, we get:
(-2)^3 - 2(-2)^2 - 3(-2) + 4 = 0
So x = -2 is not a root.

Therefore, the roots of the equation x^3 - 2x^2 - 3x + 4 = 0 are x = 1 and x = -1. We can verify this by factoring the equation as (x - 1)(x + 1)(x - 4) = 0, which shows that the roots are indeed x = 1, x = -1, and x = 4.