Answer:
D) 2 complex zeros; rational zeros among ±{1, 5, 7, 35}
Explanation:
The signs of the coefficients of the polynomial are (+ + + +). They are all the same; there are 0 sign changes. When odd-degree terms change sign, the signs are (- + - +). There are 3 changes: -→+, +→-, -→+.
Descarte's Rule of Signs tells you the lack of sign changes among the coefficients means there are 0 positive real roots. When the signs of odd-degree terms are changed, there are then 3 sign changes, indicating 1 or 3 negative real roots and 2 or 0 complex roots.
There are possibly 2 complex roots.
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The Rational Root Theorem tells you any possible rational roots will be divisors of the constant term, 35. Those divisors are ±1, ±5, ±7, ±35.
Possible rational roots are ±1, ±5, ±7, ±35.
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Additional comments
A graphing calculator shows the one rational root to be x=-5. The other two real roots are irrational (-3±√2). All are negative, as predicted.
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Complex roots always come in conjugate pairs, so there will always be an even number of complex roots. This means an odd-degree polynomial will always have at least one (1) real root.
We can shorten the list of possible rational roots by remembering that there are 0 positive real roots. Then possible rational roots are -1, -5, -7, -35. In this problem, you're merely supposed to apply the rule, and not refine the result.
Changing the signs of odd-degree terms is equivalent to reflecting the function over the y-axis. That suddenly makes negative real roots into positive real roots, so the sign change rule can be applied as for positive real roots.