The zeros of the algebraic function f(x) = 2x^3 - 11x^2 - 12x + 36 are x = 6, x = (3 + √57) / 4, and x = (3 - √57) / 4.
Given the function f(x) = 2x^3 - 11x^2 - 12x + 36 and f(6) = 0, we can find the zeros algebraically by using synthetic division. Firstly, since f(6) = 0, we can conclude that x - 6 is a factor of f(x) by the factor theorem. Performing synthetic division by dividing f(x) by x - 6 yields the quotient 2x^2 - 3x - 6.
Now, we need to factorize the quadratic expression 2x^2 - 3x - 6. We can use the quadratic formula x = (-b ± √(b² - 4ac)) / 2a to find the roots. For the quadratic, 2x² - 3x - 6, the values of a, b, and c are 2, -3, and -6 respectively.
Calculating the discriminant Δ = b² - 4ac gives Δ = (-3)² - 4(2)(-6) = 9 + 48 = 57. Since Δ > 0, the quadratic has two real and distinct roots.
Applying the quadratic formula, we get: x = (3 ± √57) / 4.
Therefore, the zeros of f(x) are x = 6, x = (3 + √57) / 4, and x = (3 - √57) / 4.