Answer:
[tex]\pm \frac{1}{4}, \pm \frac{2}{3}, \pm \frac{9}{2}[/tex]
Step-by-step explanation:
Given equation,
[tex]12x^3 + 14x^2 - x + 18=0[/tex]
By the rational root theorem,
The possible roots of a polynomial are,
[tex]\pm (\frac{\text{Factors of constant term}}{\text{Factors of leading coefficient}})[/tex]
Here, the constant term = 18 and leading coefficient = 12,
∵ Factors of 18 = 1, 2, 3, 6, 9, 18,
Factors of 12 = 1, 2, 3, 4, 6, 12,
Thus, possible roots are,
[tex]\pm (\frac{1, 2, 3, 6, 9, 18}{1, 2, 3, 4, 6, 12})[/tex]
[tex]\pm (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{12}, 2, \frac{2}{3}, 3, \frac{3}{2}, , \frac{3}{4}, 6, 9, \frac{9}{2}, \frac{9}{4}, 18)[/tex]
Hence, the correct answer are,
[tex]\pm \frac{1}{4}, \pm \frac{2}{3}, \pm \frac{9}{2}[/tex]
