Question 1117694: With the polynomial: f(x) = 5x^3 + 8x^2 -4x + 3
a. The Fundamental Theorem of Algebra states that this polynomial has ______ roots.
b. Find f(-x).
c. Use Descartes’ rule:
Number of positive real zeros
Number of negative real zeros
Number of imaginary real zeros
d. Use the Rational Root Theorem to determine the possible rational roots of f(x).
correct?: ±1, ±(1/5), ±3, ± (3/5)
e. Of the possible rational roots above, which ones are roots?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
First, your language. Although it is not a 100% hard and fast rule, it is more correct and certainly more descriptive to use the word "roots" when referring to an equation, i.e.  , and "zeros" when referring to a function, such as the example that is the subject of your question.
Then you asked for the number of "imaginary real zeros". I suspect that this was just a bit of inattention on your part, but, be that as it may, there really aren't any "imaginary" zeros either. There are complex zeros, that is zeros that are complex numbers that have a non-zero imaginary part.
The Fundamental Theorem of Algebra: Every polynomial ) of degree  has values  (some of them possibly degenerate) for which  = 0) .
To find ) , just change the sign on all terms of odd degree. Note: The constant term is degree 0, an even number.
Descarte's Rule of Signs.
In ) count the sign changes: There is a change from the second degree term to the first degree term and from the first degree term to the constant term. 2 changes. The maximum possible number of positive real zeros is 2. If it is less, the count goes down by 2. Hence, there are either 2 positive real zeros, or zero positive real zeros.
In count the sign changes: You will find 1. The maximum number of negative real zeros is 1 and the minimum number is 1 as well. One negative real zero is guaranteed.
If there are 2 positive real zeros, then all three zeros have been accounted for. If there are zero positive real zeros, then the two unaccounted for zeros must be complex. It cannot be that there is one of each because complex zeros ALWAYS appear in conjugate pairs. That is to say that if  is a zero of a polynomial function, then  must also be a zero of that polynomial function.
Your part d is correct.
Use synthetic division to test whether any of the eight possible rational zeros are, in fact, zeros of this polynomial function. Look up Synthetic Division on www.purplemath.com if you need a refresher on the synthetic division process.
John
My calculator said it, I believe it, that settles it
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