according to the rational root theorem, the following are potential roots of $f(x)=60x^{2}-57x…

Explanation:

Step1: Substitute -6/5 into f(x)

$f(-\frac{6}{5})=60\times(-\frac{6}{5})^{2}-57\times(-\frac{6}{5}) - 18=60\times\frac{36}{25}+\frac{342}{5}-18=\frac{2160}{25}+\frac{1710}{25}-\frac{450}{25}=\frac{2160 + 1710-450}{25}=\frac{3420}{25}\neq0$

Step2: Substitute -1/4 into f(x)

$f(-\frac{1}{4})=60\times(-\frac{1}{4})^{2}-57\times(-\frac{1}{4})-18=60\times\frac{1}{16}+\frac{57}{4}-18=\frac{60}{16}+\frac{228}{16}-\frac{288}{16}=\frac{60 + 228-288}{16}=0$

Step3: Substitute 3 into f(x)

$f(3)=60\times3^{2}-57\times3 - 18=60\times9-171-18=540-171 - 18=351\neq0$

Step4: Substitute 6 into f(x)

$f(6)=60\times6^{2}-57\times6-18=60\times36 - 342-18=2160-342-18=1800\neq0$

Answer:

$-\frac{1}{4}$