according to the rational root theorem, the numbers below are some of the potential roots of f(x)=10x^3 +…

Explanation:

Step1: Recall rational - root theorem

For a polynomial (a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0), the possible rational roots are of the form (\frac{p}{q}), where (p) is a factor of the constant term (a_0) and (q) is a factor of the leading - coefficient (a_n). For the polynomial (f(x)=10x^3 + 29x^2-66x + 27), (a_n = 10) and (a_0=27). The factors of (a_0 = 27) are (p=\pm1,\pm3,\pm9,\pm27), and the factors of (a_n = 10) are (q=\pm1,\pm2,\pm5,\pm10). The possible rational roots are (\frac{p}{q}=\pm1,\pm3,\pm9,\pm27,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{9}{2},\pm\frac{27}{2},\pm\frac{1}{5},\pm\frac{3}{5},\pm\frac{9}{5},\pm\frac{27}{5},\pm\frac{1}{10},\pm\frac{3}{10},\pm\frac{9}{10},\pm\frac{27}{10}).

Step2: Test the given values

Let's test the given values one by one:

• For (x = 3): (f(3)=10\times3^3+29\times3^2-66\times3 + 27=10\times27+29\times9 - 198 + 27=270+261-198 + 27=360\neq0).
• For (x = 1): (f(1)=10\times1^3+29\times1^2-66\times1 + 27=10 + 29-66 + 27=-0).
• For (x=\frac{3}{5}): (f(\frac{3}{5})=10\times(\frac{3}{5})^3+29\times(\frac{3}{5})^2-66\times\frac{3}{5}+27) (=10\times\frac{27}{125}+29\times\frac{9}{25}-\frac{198}{5}+27) (=\frac{270}{125}+\frac{261}{25}-\frac{198}{5}+27) (=\frac{270 + 1305-4950 + 3375}{125}=\frac{270+1305+3375 - 4950}{125}=0).
• For (x=\frac{9}{10}): (f(\frac{9}{10})=10\times(\frac{9}{10})^3+29\times(\frac{9}{10})^2-66\times\frac{9}{10}+27) (=10\times\frac{729}{1000}+29\times\frac{81}{100}-\frac{594}{10}+27) (=\frac{729}{100}+\frac{2349}{100}-\frac{5940}{100}+\frac{2700}{100}=\frac{729 + 2349+2700 - 5940}{100}\neq0).
• For (x = \frac{9}{2}): (f(\frac{9}{2})=10\times(\frac{9}{2})^3+29\times(\frac{9}{2})^2-66\times\frac{9}{2}+27) (=10\times\frac{729}{8}+29\times\frac{81}{4}- \frac{594}{2}+27) (=\frac{7290}{8}+\frac{2349}{4}-297 + 27) (=\frac{7290+4698-2376 + 216}{8}\neq0).

Answer:

1, (\frac{3}{5})