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

according to the rational root theorem, the following are potential roots of $f(x)=60x^{2}-57x - 18$.\n$-\frac{6}{5},-\frac{1}{4},3,6$\nwhich is an actual root of $f(x)$?\n$-\frac{6}{5}$\n$-\frac{1}{4}$\n$3$\n$6$

according to the rational root theorem, the following are potential roots of $f(x)=60x^{2}-57x - 18$.\n$-\frac{6}{5},-\frac{1}{4},3,6$\nwhich is an actual root of $f(x)$?\n$-\frac{6}{5}$\n$-\frac{1}{4}$\n$3$\n$6$

Answer

Explanation:

Step1: Substitute $x = -\frac{6}{5}$ 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 $x=-\frac{1}{4}$ 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 $x = 3$ into $f(x)$

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

Step4: Substitute $x = 6$ into $f(x)$

$f(6)=60\times6^2-57\times6-18=60\times36-342 - 18=2160-342-18=1794\neq0$

Answer:

$-\frac{1}{4}$