Answer:
[tex]\pm 1, \pm\dfrac{1}{3},\pm\dfrac{1}{9},\pm 7, \pm\dfrac{7}{3},\pm\dfrac{7}{9}[/tex].
Step-by-step explanation:
According to the Rational Root Theorem, the potential roots of a polynomial are
[tex]x=\pm\dfrac{p}{q}[/tex]
where, p is a factor of constant and q is a factor of leading term.
The given polynomial is
[tex]f(x)=9x^8+9x^6-12x+7[/tex]
Here, 9 is the leading term and 7 is constant.
Factors of 9 are ±1, ±3, ±9.
Factors of 7 are ±1, ±7.
Using rational root theorem, the rational or potential roots are
[tex]x=\pm 1, \pm\dfrac{1}{3},\pm\dfrac{1}{9},\pm 7, \pm\dfrac{7}{3},\pm\dfrac{7}{9}[/tex]
Therefore, the potential root of f(x) are [tex]\pm 1, \pm\dfrac{1}{3},\pm\dfrac{1}{9},\pm 7, \pm\dfrac{7}{3},\pm\dfrac{7}{9}[/tex].
