Question

Name the possible rational roots for the function using the Rational Root Theorem.

Answer 1

Answer:

$\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{3},\pm \frac{5}{3}$

Step-by-step explanation:

Given:

The function is given as:

$f(x)= 3x^{3}-19x^{2} +23x-15$

The possible rational roots for a function $f(x) = a_{n}x^{n}-a_{n-1}x^{n-1} +a_{n-2}x^{n-2}+....a_{1}x+a_{0}$ using the Rational Root Theorem is given as:

$\pm (\frac{\textrm{Factors of }a_{0}}{\textrm{Factors of } a_{n}}})$

Here, $a_{0}=-15,a_{n}=3$

Factors of -15 = $\pm1,\pm3,\pm5,\pm15$

Factors of 3 = $\pm1,\pm3$

Now, possible roots are given as:

$\pm (\frac{\textrm{Factors of }a_{0}}{\textrm{Factors of } a_{n}}})\\\\=\pm(\frac{\pm1,\pm3,\pm5,\pm15}{\pm1,\pm3})\\\\=\pm\frac{1}{1},\pm\frac{1}{3},\pm\frac{3}{1},\pm\frac{3}{3},\pm\frac{5}{1},\pm\frac{5}{3},\pm\frac{15}{1},\pm\frac{15}{3}\\\\=\pm1,\pm\frac{1}{3},\pm3,\pm1,\pm5,\pm\frac{5}{3},\pm15,\pm5\\\\=\pm1,\pm \frac{1}{3},\pm3, \pm 5, \pm \frac{5}{3},\pm15$

Therefore, option 1 is correct.