Polynomial Factorization

I think it must be  3x^3 - 20x^2 + kx + 12   ????

3 is one  root

3  [  3     -  20           k        12    ]

                   9        -33       3k - 99

    ____________________________

       3       -11      k - 33     3k - 87

3k - 87 =  0

3k =  87

k = 29

3  [  3      -20         29       12  ]   

                  9         -33     -12

   _________________________

      3        -11         -4         0

The remaining polynomial is

3x^2  -11x  - 4   = 0

(3x  + 1) ( x - 4)  = 0

The other two roots   are   -1/3   and   4

I assume you mean $3x^3$ and not $3x^2$, because it is impractical to write polynomials in an un-simplified way. (I hope you understand what I meant, my English is not that good)

We state the factor theorem as follows:

Let $p(x)$ be a polynomial. If $p(x)$ is divisible by $x - k$, then $p(k) = 0$.

This is a well-known result in algebra. We prove it using division algorithm: Suppose that when p(x) is divided by x - k, the quotient is q(x) and the remainder is r. (The remainder is a constant because it must have lower degree than the divisor x - k.) By division algorithm, we must have $p(x) = (x - k)q(x) + r$, substituting x = k gives $r = p(k) + 0q(k) = p(k)$. Therefore the factor theorem is equivalent to saying "If p(x) is divisible by x - k, then the remainder is 0 when p(x) is divided by by x - k", which is obviously true.

We can use this result to solve the problem. From the factor theorem, we know that $f(3) = 0$. Then, substituting x = 3 into the definition of f(x) gives $3k -87 = 0$, which gives k = 29. 

Suppose $f(x) = (x - 3)(ax^2 + bx + c)$. Then $ax^3 + (b - 3a)x^2 + (c - 3b)x - 3c = 3x^2 - 20x^2 + 29x + 12$. Comparing coefficients gives $a = 3, b = -11, c = -4$. Then we have $f(x) = (x - 3)(3x^2 - 11x - 4)$. Factorizing the quadratic part further, we have $f(x) = (x - 3)(3x + 1)(x - 4)$.

For the bonus part, simple inspection gives the roots x = 3, x = 4, and x = -1/3.