Concept explainers
State the possible number of real zeros, negative real zeros, and imaginary zeros of each function.
Answer to Problem 31PPS
Explanation of Solution
Given information:
The given equation is
The given equation has degree 6, therefore, it has 6 zeros. To find the positive real zeros, count the number of changes in sign for the coefficients of
There are 2 sign changes, so there are 2 or 0 positive real zeros.
There are 2 sign changes, so there are 2 or 0 negative real zero.
The table of possible combinations of real and imaginary zeros.
Chapter 5 Solutions
Glencoe Algebra 2 Student Edition C2014
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