according to the rational root theorem, which is a factor of the polynomial $f(x)=3x^{3}-5x^{2}-12x +…

Explanation:

Step1: Recall Rational Root Theorem

The Rational Root Theorem states that if a polynomial (a_nx^n+\cdots + a_1x + a_0) has a rational root (p/q), then (p) is a factor of the constant term (a_0) and (q) is a factor of the leading - coefficient (a_n). For the polynomial (f(x)=3x^{3}-5x^{2}-12x + 20), (a_n = 3) and (a_0=20). The possible rational roots are of the form (\pm\frac{p}{q}), where (p) is a factor of (20) (i.e., (p=\pm1,\pm2,\pm4,\pm5,\pm10,\pm20)) and (q) is a factor of (3) (i.e., (q = \pm1,\pm3)).

Step2: Test the roots for each factor

If (3x - 5=0), then (x=\frac{5}{3}). Substitute (x = \frac{5}{3}) into (f(x)): [ \begin{align*} f\left(\frac{5}{3}\right)&=3\left(\frac{5}{3}\right)^{3}-5\left(\frac{5}{3}\right)^{2}-12\left(\frac{5}{3}\right)+20\ &=3\times\frac{125}{27}-5\times\frac{25}{9}-20 + 20\ &=\frac{125}{9}-\frac{125}{9}-20 + 20\ &=0 \end{align*} ] Since (f\left(\frac{5}{3}\right)=0), by the Factor Theorem (if (f(c)=0), then (x - c) is a factor of (f(x))), and since (3x-5 = 3\left(x-\frac{5}{3}\right)), (3x - 5) is a factor of (f(x)).

Answer:

D. (3x - 5)