By using the factor theorem,show that $(x-3)$ is a factor of the polynomial $p(x) = 12x^3 - 31x^2 - 18x + 9$ and then factorise the polynomial.

(A) Step $1$: To show $(x-3)$ is a factor,we evaluate $p(3)$.
$p(3) = 12(3)^3 - 31(3)^2 - 18(3) + 9$
$p(3) = 12(27) - 31(9) - 54 + 9$
$p(3) = 324 - 279 - 54 + 9 = 0$.
Since $p(3) = 0$,by the factor theorem,$(x-3)$ is a factor of $p(x)$.
Step $2$: Divide $p(x)$ by $(x-3)$ to get the quadratic factor.
$12x^3 - 31x^2 - 18x + 9 = (x-3)(12x^2 + 5x - 3)$.
Step $3$: Factorise the quadratic $12x^2 + 5x - 3$.
$12x^2 + 9x - 4x - 3 = 3x(4x + 3) - 1(4x + 3) = (3x - 1)(4x + 3)$.
Thus,the complete factorisation is $(x-3)(3x-1)(4x+3)$.