Find all zeros of a function, 6x^4-7x^3-57x^2+63x+27 = 0.
By inspection, x = 3 is a root, hence (x-3) is a factor. Taking out this factor,
(x-3)(6x^3 + 11x^2 - 24x - 9) = 0
Again by inspection, x = -3 is a root, hence (x+3) is a factor. Taking out this factor,
(x-3)(x+3)(6x^2 - 7x - 3) = 0
And the quadratic factorises to give,
(x-3)(x+3)((2x - 3)(3x + 1) = 0
The zeros of the function then are: x = 3, x = -3, x = 3/2, x = -1/3