Answer :
We begin with the polynomial
[tex]$$
f(x) = 9x^7 + 22x^4 - 2x^3 - 6.
$$[/tex]
According to the Rational Root Theorem, any rational zero of the polynomial must be of the form
[tex]$$
\frac{p}{q},
$$[/tex]
where [tex]$p$[/tex] is a factor of the constant term and [tex]$q$[/tex] is a factor of the leading coefficient.
Step 1. Identify the factors of the constant term.
The constant term is [tex]$-6$[/tex]. Its factors (ignoring the negative sign initially) are:
[tex]$$
1, 2, 3, 6.
$$[/tex]
Taking into account both positive and negative possibilities, we have:
[tex]$$
\pm 1, \pm 2, \pm 3, \pm 6.
$$[/tex]
Step 2. Identify the factors of the leading coefficient.
The leading coefficient is [tex]$9$[/tex]. Its factors are:
[tex]$$
1, 3, 9.
$$[/tex]
Again, we consider both positive and negative versions, so:
[tex]$$
\pm 1, \pm 3, \pm 9.
$$[/tex]
Step 3. Form the possible rational zeros.
Each possible rational zero is of the form
[tex]$$
\frac{p}{q},
$$[/tex]
where [tex]$p$[/tex] is taken from [tex]$\{\pm 1, \pm 2, \pm 3, \pm 6\}$[/tex] and [tex]$q$[/tex] is taken from [tex]$\{\pm 1, \pm 3, \pm 9\}$[/tex].
This gives us the following distinct possibilities:
1. Using [tex]$q = 1$[/tex]:
[tex]$$
\pm \frac{1}{1} = \pm 1, \quad \pm \frac{2}{1} = \pm 2, \quad \pm \frac{3}{1} = \pm 3, \quad \pm \frac{6}{1} = \pm 6.
$$[/tex]
2. Using [tex]$q = 3$[/tex]:
[tex]$$
\pm \frac{1}{3}, \quad \pm \frac{2}{3}, \quad \pm \frac{3}{3} = \pm 1, \quad \pm \frac{6}{3} = \pm 2.
$$[/tex]
(Note that [tex]$\pm 1$[/tex] and [tex]$\pm 2$[/tex] are already in the list.)
3. Using [tex]$q = 9$[/tex]:
[tex]$$
\pm \frac{1}{9}, \quad \pm \frac{2}{9}, \quad \pm \frac{3}{9} = \pm \frac{1}{3}, \quad \pm \frac{6}{9} = \pm \frac{2}{3}.
$$[/tex]
(Again, [tex]$\pm \frac{1}{3}$[/tex] and [tex]$\pm \frac{2}{3}$[/tex] are already listed.)
Step 4. Write the complete list of all distinct possible rational zeros.
Thus, the possible rational zeros of the polynomial are:
[tex]$$
\pm 1, \;\pm 2, \;\pm 3, \;\pm 6, \;\pm \frac{1}{3}, \;\pm \frac{2}{3}, \;\pm \frac{1}{9}, \;\pm \frac{2}{9}.
$$[/tex]
This is the full set of candidates based on the Rational Root Theorem.
[tex]$$
f(x) = 9x^7 + 22x^4 - 2x^3 - 6.
$$[/tex]
According to the Rational Root Theorem, any rational zero of the polynomial must be of the form
[tex]$$
\frac{p}{q},
$$[/tex]
where [tex]$p$[/tex] is a factor of the constant term and [tex]$q$[/tex] is a factor of the leading coefficient.
Step 1. Identify the factors of the constant term.
The constant term is [tex]$-6$[/tex]. Its factors (ignoring the negative sign initially) are:
[tex]$$
1, 2, 3, 6.
$$[/tex]
Taking into account both positive and negative possibilities, we have:
[tex]$$
\pm 1, \pm 2, \pm 3, \pm 6.
$$[/tex]
Step 2. Identify the factors of the leading coefficient.
The leading coefficient is [tex]$9$[/tex]. Its factors are:
[tex]$$
1, 3, 9.
$$[/tex]
Again, we consider both positive and negative versions, so:
[tex]$$
\pm 1, \pm 3, \pm 9.
$$[/tex]
Step 3. Form the possible rational zeros.
Each possible rational zero is of the form
[tex]$$
\frac{p}{q},
$$[/tex]
where [tex]$p$[/tex] is taken from [tex]$\{\pm 1, \pm 2, \pm 3, \pm 6\}$[/tex] and [tex]$q$[/tex] is taken from [tex]$\{\pm 1, \pm 3, \pm 9\}$[/tex].
This gives us the following distinct possibilities:
1. Using [tex]$q = 1$[/tex]:
[tex]$$
\pm \frac{1}{1} = \pm 1, \quad \pm \frac{2}{1} = \pm 2, \quad \pm \frac{3}{1} = \pm 3, \quad \pm \frac{6}{1} = \pm 6.
$$[/tex]
2. Using [tex]$q = 3$[/tex]:
[tex]$$
\pm \frac{1}{3}, \quad \pm \frac{2}{3}, \quad \pm \frac{3}{3} = \pm 1, \quad \pm \frac{6}{3} = \pm 2.
$$[/tex]
(Note that [tex]$\pm 1$[/tex] and [tex]$\pm 2$[/tex] are already in the list.)
3. Using [tex]$q = 9$[/tex]:
[tex]$$
\pm \frac{1}{9}, \quad \pm \frac{2}{9}, \quad \pm \frac{3}{9} = \pm \frac{1}{3}, \quad \pm \frac{6}{9} = \pm \frac{2}{3}.
$$[/tex]
(Again, [tex]$\pm \frac{1}{3}$[/tex] and [tex]$\pm \frac{2}{3}$[/tex] are already listed.)
Step 4. Write the complete list of all distinct possible rational zeros.
Thus, the possible rational zeros of the polynomial are:
[tex]$$
\pm 1, \;\pm 2, \;\pm 3, \;\pm 6, \;\pm \frac{1}{3}, \;\pm \frac{2}{3}, \;\pm \frac{1}{9}, \;\pm \frac{2}{9}.
$$[/tex]
This is the full set of candidates based on the Rational Root Theorem.
