Answer:
f(x) = 3(x + 2)(x - 1)(x + 3)
Explanation:
A logical first step would be to factor 3 out of all four terms:
f(x) = 3x^3+12x^2+3x-18 = 3(x^3 + 4x^2 + x - 6)
Roots of this x^3 + 4x^2 + x - 6 could be factors of 6: {±1, ±2, ±3, ±6}.
I would use synthetic division here to determine which, if any, of these possibilities are actually roots of x^3 + 4x^2 + x - 6. Let's try x = 1 and see whether the remainder of this synth. div. is 0, which would indicate that 1 is indeed a root of x^3 + 4x^2 + x - 6:
1 / 1 4 1 -6
1 5 6
------------------------
1 5 6 0
Yes, 1 is a root of x^3 + 4x^2 + x - 6, and so (x - 1) is a factor of x^3 + 4x^2 + x - 6.
Look at the coefficients of the quotient, which are 1, 5 and 6.
This represents the quadratic 1x² + 5x + 6, whose factors are (x + 2) and
(x + 3).
Thus, the given polynomial in factored form is:
f(x) = 3x^3+12x^2+3x-18 = 3(x + 2)(x - 1)(x + 3)