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Answer :
The total number of real zeros of [tex]\( f(x) \)[/tex] is 4.
- Given Factors
- The function [tex]\( f(x) \) has \( (x - 1) \) and \( (x + 3i) \)[/tex] as factors.
- Since [tex]\( (x + 3i) \)[/tex] is a factor and the coefficients of [tex]\( f(x) \)[/tex] are real, the complex conjugate [tex]\( (x - 3i) \)[/tex] must also be a factor.
- Form the Factors
- The factors are: [tex]\( (x - 1) \), \( (x + 3i) \), and \( (x - 3i) \)[/tex].
- Polynomial and Degree
- [tex]\( f(x) \)[/tex] is a polynomial of degree 6, so it has 6 roots in total, counting multiplicities.
- We have identified 3 factors so far, corresponding to 3 roots:
- [tex]\( x = 1 \)[/tex] (real root)
- [tex]\( x = -3i \)[/tex] (complex root)
- [tex]\( x = 3i \)[/tex] (complex root)
- Remaining Roots
- Since [tex]\( f(x) \)[/tex] has 6 roots in total and we have identified 3, there are 3 remaining roots.
- We need to determine if these remaining roots are real or complex.
- Descartes' Rule of Signs
- To determine the number of real zeros, we can apply Descartes' Rule of Signs:
- For [tex]\( f(x) \)[/tex]:
- The signs of the coefficients are [tex]\( +, -, +, -, +, -, - \)[/tex].
- There are 6 sign changes, indicating up to 6 positive real zeros or fewer (by even number decrements).
- For [tex]\( f(-x) \)[/tex]:
- Substitute [tex]\(-x\) into \( f(x) \): \( 3(-x)^6 - 17(-x)^5 + 39(-x)^4 - 45(-x)^3 + 26(-x)^2 - 6(-x) = 3x^6 + 17x^5 + 39x^4 + 45x^3 + 26x^2 + 6x \)[/tex]
- The signs of the coefficients are [tex]\( +, +, +, +, +, +, + \)[/tex], indicating 0 sign changes, so there are 0 negative real zeros.
- Conclusion
- Given that there are 0 negative real zeros and that 3 of the roots are accounted for by the identified factors (1 real, 2 complex), the remaining 3 roots must all be real.
- Hence, the total number of real zeros is [tex]\( 1 (from x = 1) + 3 (additional real roots) = 4 \)[/tex]
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