Answer :
To find all of the possible rational zeros of the polynomial [tex]\( f(x) = -3x^4 - 9x^3 - 6x^2 - 8x + 14 \)[/tex], we use the Rational Root Theorem. According to this theorem, any possible rational zero of the polynomial can be expressed as [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term (14 in this case) and [tex]\( q \)[/tex] is a factor of the leading coefficient (-3 in this case).
Step-by-Step Solution:
1. Identify the constant term and its factors:
The constant term [tex]\( a_0 = 14 \)[/tex].
Factors of 14 are: [tex]\( \pm 1, \pm 2, \pm 7, \pm 14 \)[/tex].
2. Identify the leading coefficient and its factors:
The leading coefficient [tex]\( a_n = -3 \)[/tex].
Factors of -3 are: [tex]\( \pm 1, \pm 3 \)[/tex].
3. Form possible rational zeros [tex]\( \frac{p}{q} \)[/tex]:
We consider [tex]\( \frac{\text{factors of } 14}{\text{factors of } -3} \)[/tex].
This gives us the following possible rational zeros:
- Using factors of 14: [tex]\( \pm 1, \pm 2, \pm 7, \pm 14 \)[/tex]
- Using factors of -3: [tex]\( \frac{\pm 1}{\pm 1}, \frac{\pm 2}{\pm 1}, \frac{\pm 7}{\pm 1}, \frac{\pm 14}{\pm 1}, \frac{\pm 1}{\pm 3}, \frac{\pm 2}{\pm 3}, \frac{\pm 7}{\pm 3}, \frac{\pm 14}{\pm 3} \)[/tex]
4. Simplify to get unique candidates for rational zeros:
- [tex]\( \frac{\pm 1}{\pm 1} = \pm 1 \)[/tex]
- [tex]\( \frac{\pm 2}{\pm 1} = \pm 2 \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 1} = \pm 7 \)[/tex]
- [tex]\( \frac{\pm 14}{\pm 1} = \pm 14 \)[/tex]
- [tex]\( \frac{\pm 1}{\pm 3} = \pm \frac{1}{3} \)[/tex]
- [tex]\( \frac{\pm 2}{\pm 3} = \pm \frac{2}{3} \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 3} = \pm \frac{7}{3} \)[/tex]
- [tex]\( \frac{\pm 14}{\pm 3} = \pm \frac{14}{3} \)[/tex]
Thus, combining all these, the possible rational zeros of the polynomial [tex]\( f(x) = -3x^4 - 9x^3 - 6x^2 - 8x + 14 \)[/tex] include:
[tex]\[ \pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3} \][/tex]
Given our options:
- [tex]\(\pm 1, \pm 2, \pm 7, \pm 14\)[/tex]
- [tex]\(\pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}\)[/tex]
- [tex]\(\pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}\)[/tex]
- [tex]\(\pm 1, \pm 2, \pm 7, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}\)[/tex]
The correct list is:
[tex]\[ \pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3} \][/tex]
Step-by-Step Solution:
1. Identify the constant term and its factors:
The constant term [tex]\( a_0 = 14 \)[/tex].
Factors of 14 are: [tex]\( \pm 1, \pm 2, \pm 7, \pm 14 \)[/tex].
2. Identify the leading coefficient and its factors:
The leading coefficient [tex]\( a_n = -3 \)[/tex].
Factors of -3 are: [tex]\( \pm 1, \pm 3 \)[/tex].
3. Form possible rational zeros [tex]\( \frac{p}{q} \)[/tex]:
We consider [tex]\( \frac{\text{factors of } 14}{\text{factors of } -3} \)[/tex].
This gives us the following possible rational zeros:
- Using factors of 14: [tex]\( \pm 1, \pm 2, \pm 7, \pm 14 \)[/tex]
- Using factors of -3: [tex]\( \frac{\pm 1}{\pm 1}, \frac{\pm 2}{\pm 1}, \frac{\pm 7}{\pm 1}, \frac{\pm 14}{\pm 1}, \frac{\pm 1}{\pm 3}, \frac{\pm 2}{\pm 3}, \frac{\pm 7}{\pm 3}, \frac{\pm 14}{\pm 3} \)[/tex]
4. Simplify to get unique candidates for rational zeros:
- [tex]\( \frac{\pm 1}{\pm 1} = \pm 1 \)[/tex]
- [tex]\( \frac{\pm 2}{\pm 1} = \pm 2 \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 1} = \pm 7 \)[/tex]
- [tex]\( \frac{\pm 14}{\pm 1} = \pm 14 \)[/tex]
- [tex]\( \frac{\pm 1}{\pm 3} = \pm \frac{1}{3} \)[/tex]
- [tex]\( \frac{\pm 2}{\pm 3} = \pm \frac{2}{3} \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 3} = \pm \frac{7}{3} \)[/tex]
- [tex]\( \frac{\pm 14}{\pm 3} = \pm \frac{14}{3} \)[/tex]
Thus, combining all these, the possible rational zeros of the polynomial [tex]\( f(x) = -3x^4 - 9x^3 - 6x^2 - 8x + 14 \)[/tex] include:
[tex]\[ \pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3} \][/tex]
Given our options:
- [tex]\(\pm 1, \pm 2, \pm 7, \pm 14\)[/tex]
- [tex]\(\pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}\)[/tex]
- [tex]\(\pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}\)[/tex]
- [tex]\(\pm 1, \pm 2, \pm 7, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}\)[/tex]
The correct list is:
[tex]\[ \pm 1, \pm 2, \pm 7, \pm 14, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3} \][/tex]
