Answer:
[tex]\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{3},\pm \frac{5}{3}[/tex]
Step-by-step explanation:
Given:
The function is given as:
[tex]f(x)= 3x^{3}-19x^{2} +23x-15[/tex]
The possible rational roots for a function [tex]f(x) = a_{n}x^{n}-a_{n-1}x^{n-1} +a_{n-2}x^{n-2}+....a_{1}x+a_{0}[/tex] using the Rational Root Theorem is given as:
[tex]\pm (\frac{\textrm{Factors of }a_{0}}{\textrm{Factors of } a_{n}}})[/tex]
Here, [tex]a_{0}=-15,a_{n}=3[/tex]
Factors of -15 = [tex]\pm1,\pm3,\pm5,\pm15[/tex]
Factors of 3 = [tex]\pm1,\pm3[/tex]
Now, possible roots are given as:
[tex]\pm (\frac{\textrm{Factors of }a_{0}}{\textrm{Factors of } a_{n}}})\\\\=\pm(\frac{\pm1,\pm3,\pm5,\pm15}{\pm1,\pm3})\\\\=\pm\frac{1}{1},\pm\frac{1}{3},\pm\frac{3}{1},\pm\frac{3}{3},\pm\frac{5}{1},\pm\frac{5}{3},\pm\frac{15}{1},\pm\frac{15}{3}\\\\=\pm1,\pm\frac{1}{3},\pm3,\pm1,\pm5,\pm\frac{5}{3},\pm15,\pm5\\\\=\pm1,\pm \frac{1}{3},\pm3, \pm 5, \pm \frac{5}{3},\pm15[/tex]
Therefore, option 1 is correct.
