To determine the possible rational zeros for the polynomial function [tex]\( f(x) = x^5 - 6x^2 + 2x + 3 \)[/tex] using the Rational Zero Theorem, we follow these steps:
The Rational Zero Theorem states that any potential rational zero of the polynomial function [tex]\( f(x) \)[/tex] in the form of [tex]\( \frac{p}{q} \)[/tex] must be such that:
- [tex]\( p \)[/tex] is a factor of the constant term of the polynomial.
- [tex]\( q \)[/tex] is a factor of the leading coefficient of the polynomial.
For our polynomial [tex]\( f(x) = x^5 - 6x^2 + 2x + 3 \)[/tex] :
- The constant term (the term without [tex]\( x \)[/tex]) is [tex]\( 3 \)[/tex].
- The leading coefficient (the coefficient of the highest power of [tex]\( x \)[/tex]) is [tex]\( 1 \)[/tex].
1. Identify factors of the constant term ([tex]\( p \)[/tex]):
- The factors of [tex]\( 3 \)[/tex] are [tex]\( \pm 1, \pm 3 \)[/tex].
2. Identify factors of the leading coefficient ([tex]\( q \)[/tex]):
- The factors of [tex]\( 1 \)[/tex] are [tex]\( \pm 1 \)[/tex].
3. Form all possible rational zeros by taking each factor of the constant term and dividing it by each factor of the leading coefficient:
- Possible rational zeros are [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p = \pm 1, \pm 3 \)[/tex] and [tex]\( q = \pm 1 \)[/tex].
4. Listing all combinations, we get:
- [tex]\( \frac{1}{1} = 1 \)[/tex]
- [tex]\( \frac{-1}{1} = -1 \)[/tex]
- [tex]\( \frac{3}{1} = 3 \)[/tex]
- [tex]\( \frac{-3}{1} = -3 \)[/tex]
Thus, the possible rational zeros are [tex]\( \pm 1, \pm 3 \)[/tex].
Therefore, the correct answer is:
D. [tex]\( \pm 1, \pm 3 \)[/tex]
