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Given the polynomial function:
[tex]\[ f(x) = 6x^3 - 61x^2 + 65x - 18 \][/tex]
Let's solve this step-by-step:
### a) List all possible rational zeros
According to the Rational Root Theorem, the possible rational zeros of a polynomial [tex]\( f(x) \)[/tex] are the ratios [tex]\(\pm \frac{p}{q}\)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term (the last term) and [tex]\( q \)[/tex] is a factor of the leading coefficient (the coefficient of the highest degree term).
1. The constant term of [tex]\( f(x) \)[/tex] is [tex]\(-18\)[/tex]. The factors of [tex]\(-18\)[/tex] are:
[tex]\(\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \)[/tex].
2. The leading coefficient of [tex]\( f(x) \)[/tex] is [tex]\( 6 \)[/tex]. The factors of [tex]\( 6 \)[/tex] are:
[tex]\(\pm 1, \pm 2, \pm 3, \pm 6 \)[/tex].
Now, list all possible ratios [tex]\(\pm \frac{p}{q}\)[/tex]:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm \frac{3}{2}, \pm 3, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18 \][/tex]
So, the answer is:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm \frac{3}{2}, \pm 3, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18 \][/tex]
### b) Find all rational zeros
The rational zeros found for [tex]\( f(x) \)[/tex] are [tex]\( \frac{1}{2} \)[/tex] and [tex]\( 9 \)[/tex].
So, the answer is [tex]\( \boxed{0.5 \text{ and } 9} \)[/tex] (equivalently [tex]\( \frac{1}{2} \)[/tex] and 9).
### c) Factor [tex]\( f(x) \)[/tex]
Given the rational roots, we can factorize the polynomial:
Using the rational zeros [tex]\( \frac{1}{2} \)[/tex] and [tex]\( 9 \)[/tex], the polynomial can be factored as:
[tex]\[ f(x) = 6 \cdot (x - 9) \cdot (2x - 1) \cdot (3x - 2) \][/tex]
Written as polynomial factors, it is:
[tex]\[ f(x) = (x - 9)(2x - 1)(3x - 2) \][/tex]
So, the final factorization of [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{(x - 9)(2x - 1)(3x - 2)} \][/tex]
### Conclusion
- Possible rational zeros: [tex]\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm \frac{3}{2}, \pm 3, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18 \)[/tex].
- Rational zeros: [tex]\( \frac{1}{2} \)[/tex] and [tex]\( 9 \)[/tex].
- Polynomial factorization: [tex]\( (x - 9)(2x - 1)(3x - 2) \)[/tex].
Among the provided answer choices, the correct ones based on the questions are:
(b) Determine the rational zeros for [tex]\( f(x) \)[/tex]:
Answer: [tex]\( \boxed{B. \; 9, \frac{1}{2}, \frac{2}{3}} \)[/tex]
(although note that [tex]\(2/3\)[/tex] is included in the answer, the actual zeros known from given are [tex]\(9\)[/tex] and [tex]\(1/2\)[/tex]).
[tex]\[ f(x) = 6x^3 - 61x^2 + 65x - 18 \][/tex]
Let's solve this step-by-step:
### a) List all possible rational zeros
According to the Rational Root Theorem, the possible rational zeros of a polynomial [tex]\( f(x) \)[/tex] are the ratios [tex]\(\pm \frac{p}{q}\)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term (the last term) and [tex]\( q \)[/tex] is a factor of the leading coefficient (the coefficient of the highest degree term).
1. The constant term of [tex]\( f(x) \)[/tex] is [tex]\(-18\)[/tex]. The factors of [tex]\(-18\)[/tex] are:
[tex]\(\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \)[/tex].
2. The leading coefficient of [tex]\( f(x) \)[/tex] is [tex]\( 6 \)[/tex]. The factors of [tex]\( 6 \)[/tex] are:
[tex]\(\pm 1, \pm 2, \pm 3, \pm 6 \)[/tex].
Now, list all possible ratios [tex]\(\pm \frac{p}{q}\)[/tex]:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm \frac{3}{2}, \pm 3, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18 \][/tex]
So, the answer is:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm \frac{3}{2}, \pm 3, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18 \][/tex]
### b) Find all rational zeros
The rational zeros found for [tex]\( f(x) \)[/tex] are [tex]\( \frac{1}{2} \)[/tex] and [tex]\( 9 \)[/tex].
So, the answer is [tex]\( \boxed{0.5 \text{ and } 9} \)[/tex] (equivalently [tex]\( \frac{1}{2} \)[/tex] and 9).
### c) Factor [tex]\( f(x) \)[/tex]
Given the rational roots, we can factorize the polynomial:
Using the rational zeros [tex]\( \frac{1}{2} \)[/tex] and [tex]\( 9 \)[/tex], the polynomial can be factored as:
[tex]\[ f(x) = 6 \cdot (x - 9) \cdot (2x - 1) \cdot (3x - 2) \][/tex]
Written as polynomial factors, it is:
[tex]\[ f(x) = (x - 9)(2x - 1)(3x - 2) \][/tex]
So, the final factorization of [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{(x - 9)(2x - 1)(3x - 2)} \][/tex]
### Conclusion
- Possible rational zeros: [tex]\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm \frac{3}{2}, \pm 3, \pm 6, \pm 9, \pm \frac{9}{2}, \pm 18 \)[/tex].
- Rational zeros: [tex]\( \frac{1}{2} \)[/tex] and [tex]\( 9 \)[/tex].
- Polynomial factorization: [tex]\( (x - 9)(2x - 1)(3x - 2) \)[/tex].
Among the provided answer choices, the correct ones based on the questions are:
(b) Determine the rational zeros for [tex]\( f(x) \)[/tex]:
Answer: [tex]\( \boxed{B. \; 9, \frac{1}{2}, \frac{2}{3}} \)[/tex]
(although note that [tex]\(2/3\)[/tex] is included in the answer, the actual zeros known from given are [tex]\(9\)[/tex] and [tex]\(1/2\)[/tex]).
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