Question

according to the rational root theorem, which number is a potential root of $f(x) = 9x^8 + 9x^6 - 12x + 7$?
○ 0
○ $\frac{2}{7}$
○ 2
○ $\frac{7}{3}$

Explanation:

Step1: Recall Rational Root Theorem

The Rational Root Theorem states that any potential rational root \(\frac{p}{q}\) of a polynomial \(a_nx^n + a_{n - 1}x^{n - 1}+\dots+a_1x + a_0\) (where \(a_n
eq0\)) is 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 \(f(x)=9x^8 + 9x^6-12x + 7\), the constant term \(a_0 = 7\) and the leading coefficient \(a_n=9\).

Step2: Find factors of \(a_0\) and \(a_n\)

• Factors of \(a_0 = 7\) (constant term) are \(\pm1,\pm7\).
• Factors of \(a_n = 9\) (leading coefficient) are \(\pm1,\pm3,\pm9\).

Step3: Determine potential rational roots

So, the potential rational roots \(\frac{p}{q}\) are of the form \(\frac{\pm1}{\pm1},\frac{\pm1}{\pm3},\frac{\pm1}{\pm9},\frac{\pm7}{\pm1},\frac{\pm7}{\pm3},\frac{\pm7}{\pm9}\). Simplifying these, we get \(\pm1,\pm\frac{1}{3},\pm\frac{1}{9},\pm7,\pm\frac{7}{3},\pm\frac{7}{9}\).

Now let's check the options:

• Option 0: \(0\) is not in the form \(\frac{p}{q}\) as per the theorem (since \(p\) would have to be a factor of \(7\) and \(q\) a factor of \(9\), and \(0\) can't be written as such with non - zero \(p\) and \(q\) from the factors we found).
• Option \(\frac{2}{7}\): The numerator \(2\) is not a factor of \(7\) (the constant term), so this is not a potential root.
• Option 2: The numerator \(2\) is not a factor of \(7\) (the constant term), so this is not a potential root.
• Option \(\frac{7}{3}\): The numerator \(7\) is a factor of \(7\) (the constant term) and the denominator \(3\) is a factor of \(9\) (the leading coefficient), so \(\frac{7}{3}\) is a potential root.

Answer:

\(\frac{7}{3}\) (the option \(\frac{7}{3}\))